Derivation of the determinant as a natural transformation on commutative rings

Question

Let $K$ be a commutative ring and $K^*$ the group of invertible elements of K.

The discussion here suggests that for $K = \mathbb{R}$, any natural transformation from $\mathrm{GL}_n(\mathbb{R})$ to $\mathbb{R}^\times$ would be $\det$ up to composition with an automorphism of $\mathbb{R^\times}$.

By trying to construct a natural transformation $\mathrm{GL}_n(K) \rightarrow K^*$, is it possible to canonically arrive at an explicit formula for the determinant?

By canonical, I mean not making a priori choices of other mathematical structures in the construction, like for instance an exterior algebra.

Answer 1

Recall that the group scheme $GL_n(-)$ can be described abstractly as the group-valued functor $R \mapsto GL_n(R)$ sending a commutative ring $R$ to the group of invertible $n \times n$ matrices over $R$. This functor happens to be affine, meaning that $GL_n(-)$ is actually represented by maps out of a commutative ring (which inherits the structure of a Hopf algebra). Explicitly, this commutative ring is given by

$$\mathcal{O}_{GL_n} \cong \mathbb{Z}[x_{ij} : 1 \le i, j \le n][\det(X)^{-1}]$$

where $X$ is the square matrix with coefficients $x_{ij}$; the easiest proof of this uses the fact that a matrix is invertible iff its determinant is invertible, but the point is that by the Yoneda lemma, once you have any way of showing that $GL_n(-)$ is affine, it is represented by a commutative ring which is unique up to unique isomorphism, so the determinant is there whether you know it's there or not.

The easiest case of this to see is the case $n = 1$: here $GL_1(-) \cong \mathbb{G}_m(-)$ is called the multiplicative group scheme, it sends a ring $R$ to its group $R^{\times}$ of units, and it's clearly represented by the commutative ring

$$\mathcal{O}_{\mathbb{G}_m} \cong \mathbb{Z}[x, x^{-1}]$$

with comultiplication $\Delta x = x \otimes x$.

Now, a natural transformation $GL_n(-) \to GL_1(-)$ of group schemes is the same thing, by the Yoneda lemma, as a morphism of Hopf algebras

$$f : \mathcal{O}_{GL_1} \cong \mathbb{Z}[x, x^{-1}] \to \mathcal{O}_{GL_n}.$$

In particular such a morphism has the property that $f(x)$ is invertible; in addition $f(x)$ must satisfy $\Delta f(x) = f(x) \otimes f(x)$ (it must be what is called a grouplike element); this is necessary and sufficient. Now, it's not hard to check that the only invertible elements of $\mathcal{O}_{GL_n}$ are of the form $\pm \det(X)^n, n \in \mathbb{Z}$, for example via a degree argument. Moreover, the comultiplication takes the form

$$\Delta (\pm \det(X)^n) = \pm \det(X)^n \otimes \det(X)^n$$

(this follows from the multiplicativity of the determinant; I recognize that I am assuming facts you might not want to assume but I'll arrive at an alternate definition of the determinant in the end), and so in fact the grouplike elements are precisely those of the form $\det(X)^n, n \in \mathbb{Z}$. The conclusion is:

Proposition: Every morphism $GL_n \to GL_1$ of group schemes is an integer power of the determinant.

Because $\mathbb{G}_m$ is commutative, the set of such morphisms is an abelian group, and so we can now define the determinant (up to a mild ambiguity) as one of the two possible generators of this group; the determinant is distinguished by the fact that it is also well-defined on not-necessarily-invertible matrices, which corresponds to the subalgebra $\mathcal{O}_{M_n} \cong \mathbb{Z}[x_{ij}]$ of $\mathcal{O}_{GL_n}$.

There's an analogous but harder version of this discussion where we use $\mathcal{O}_{M_n}$ and $\mathcal{O}_{M_1}$ instead; the conclusion will be that every morphism $M_n(-) \to M_1(-)$ of monoid schemes is a nonnegative integer power of the determinant, or equivalently that the grouplike elements of $\mathbb{Z}[x_{ij}]$ are nonnegative integer powers of $\det(X)$, so now we can define the determinant, with no ambiguity, as the unique generator of this monoid.

Answer 2

The below is in a very similar vein to Qiaochu's answer, but differs in its precise technology. I have tried to be overly explicit; due to limits of space, I only treat below the matrices-not-assumed-invertible scenario. The case of invertible matrices can be handled very similarly.

As does Qiaochu, I will presuppose some basic properties of the determinant; the point is not to derive the determinant "from scratch", but merely to give it natural post hoc justification.

What follows is the main argument. The appendix, separately posted due to the character limit, can be found here.

Henceforth fix a positive natural $\text{d}$ (the $\text{d}=0$ case is trivial but sadly also degenerate); we will specifically consider determinants of $\text{d}\times\text{d}$ matrices.

In the interest of notational compactness, we abbreviate "$\in$" as "$\colon$" and identify the notation of a natural $\text{n}$ with the set $\left\{0,\dots,\text{n}-1\right\}$.

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The actual argument:

In this section, denote by $\textbf{CRing}$ the category of commutative rings and by $\textbf{Set}$ the category of sets.

Proposition/Definition 0.: There are evident, well-defined functors

• $\mathbb{1}\ \colon\ \textbf{CRing}\to\textbf{Set}$

• $\text{EL}\ \colon\ \textbf{CRing}\to\textbf{Set}$

• $\text{DIAG}_{\text{d}}\ \colon\ \textbf{CRing}\to\textbf{Set}$

• $\text{MAT}_{\text{d},\text{d}}\ \colon\ \textbf{CRing}\to\textbf{Set}$

that map respectively by sending crings $R$

• constantly to the singleton $\mathbb{1}\left(R\right)=\left\{*\right\}$ (and cring morphisms to its identity).

• forgetfully to their underlying set $\text{EL}\left(R\right)$ of elements (and cring morphisms to their underlying functions).

• to the set $\text{DIAG}_{\text{d}}\left(R\right)$ of diagonal $\text{d}\times\text{d}$ matrices with entries therein (and cring morphisms to the functions between these sets given by acting homomorphically on entries).

• to the set $\text{MAT}_{\text{d},\text{d}}\left(R\right)$ of $\text{d}\times\text{d}$ matrices with entries therein (and cring morphisms to the functions between these sets given by acting homomorphically on entries).

There are moreover evident, well-defined natural transformations

• $\text{Id}\ \colon\ \mathbb{1} \to \text{EL}$

• $\text{Mul}_{\text{EL}}\ \colon\ \text{EL}\times \text{EL} \to \text{EL}$

• $\text{Id}_{\text{DIAG}_{\text{d}}}\ \colon\ \mathbb{1}\to\text{DIAG}_{\text{d}}$

• $\text{Mul}_{\text{DIAG}_{\text{d}}}\ \colon\ \text{DIAG}_{\text{d}}\times \text{DIAG}_{\text{d}} \to \text{DIAG}_{\text{d}}$

• $\text{Id}_{\text{MAT}_{\text{d},\text{d}}}\ \colon\ \mathbb{1}\to\text{MAT}_{\text{d},\text{d}}$

• $\text{Mul}_{\text{MAT}_{\text{d},\text{d}}}\ \colon\ \text{MAT}_{\text{d},\text{d}}\times \text{MAT}_{\text{d},\text{d}} \to \text{MAT}_{\text{d},\text{d}}$

• $\text{Det}_{\text{d}}\ \colon\ \text{MAT}_{\text{d},\text{d}}\to\text{EL}$

• $\text{Inc}'_{\text{d}}\ \colon\ \text{EL}\to\text{DIAG}_{\text{d}}$

• $\text{Inc}''_{\text{d}}\ \colon\ \text{DIAG}_{\text{d}}\to\text{MAT}_{\text{d},\text{d}}$

that map respectively at (the component at) cring $R$

• by sending the element $*\ \colon\ \mathbb{1}\left(R\right)$ of the singleton to the multiplicative identity $\text{Id}_{R}\left(*\right)\ \colon\ \text{EL}\left(R\right)$.

• by sending the pair of elements $\left(r_{0},r_{1}\right)\ \colon\ \text{EL}\left(R\right)\times \text{El}\left(R\right)$ to its product $\text{Mul}_{\text{EL},R}\left(r_{0},r_{1}\right)=r_{0}r_{1}\ \colon\ \text{EL}\left(R\right)$.

• by sending the element $*\ \colon\ \mathbb{1}\left(R\right)$ of the singleton to the (diagonal) $\text{d}\times\text{d}$ identity matrix $\text{Id}_{\text{DIAG}_{\text{d}},R}\left(*\right)\ \colon\ \text{DIAG}_{\text{d}}\left(R\right)$.

• by sending the pair of diagonal $\text{d}\times\text{d}$ matrices $\left(A_{0},A_{1}\right)\ \colon\ \text{DIAG}_{\text{d}}\left(R\right)\times \text{DIAG}_{\text{d}}\left(R\right)$ to its product $\text{Mul}_{\text{DIAG}_{\text{d}}, R}\left(A_{0},A_{1}\right)=A_{1}A_{0}\ \colon\ \text{DIAG}_{\text{d}}\left(R\right)$.

• by sending the element $*\ \colon\ \mathbb{1}\left(R\right)$ of the singleton to the $\text{d}\times\text{d}$ identity matrix $\text{Id}_{\text{MAT}_{\text{d},\text{d}},R}\left(*\right)\ \colon\ \text{MAT}_{\text{d},\text{d}}\left(R\right)$.

• by sending the pair of $\text{d}\times\text{d}$ matrices $\left(A_{0},A_{1}\right)\ \colon\ \text{MAT}_{\text{d},\text{d}}\left(R\right)\times \text{MAT}_{\text{d},\text{d}}\left(R\right)$ to its product $\text{Mul}_{\text{MAT}_{\text{d},\text{d}}, R}\left(A_{0},A_{1}\right)=A_{1}A_{0}\ \colon\ \text{MAT}_{\text{d},\text{d}}\left(R\right)$.

• by sending the matrix $A\ \colon\ \text{MAT}_{\text{d},\text{d}}\left(R\right)$ to its determinant $\text{Det}_{\text{d},R}\left(A\right)\ \colon\ \text{EL}\left(R\right)$

• by sending the element $r\ \colon\ \text{EL}\left(R\right)$ to the diagonal matrix $\text{inc}'_{\text{d},R}\left(r\right)\ \colon\ \text{DIAG}_{\text{d}}\left(R\right)$ whose upper-left-most diagonal entry is $r$ and whose remaining diagonal entries are $1$.

• by sending the $\text{d}\times\text{d}$ diagonal matrix $A\ \colon\ \text{DIAG}_{\text{d}}\left(R\right)$ to the same $\text{d}\times\text{d}$ matrix $\text{inc}''_{\text{d},R}\left(A\right)\ \colon\ \text{DIAG}_{\text{d}}\left(R\right)$.

Define also the natural transformation $\text{Inc}_{\text{d}} := \text{Inc}''_{\text{d}}\circ \text{Inc}'_{\text{d}}$ and remark that $\text{Det}_{\text{d}}\circ\text{Inc}_{\text{d}}=\text{id}_{\text{EL}}$.

"Justification" 0.: It's not hard—but certainly tedious—to check. $\Box$

Proposition 1.: We have internal monoids in the copresheaf category $\textbf{CRing}\to\textbf{Set}$ with respect to its Cartesian monoidal structure

• $\underline{\text{EL}}:=\left(\text{EL},\text{Mul}_{\text{EL}},\text{Id}_{\text{EL}}\right)$.

• $\underline{\text{DIAG}_{\text{d}}}:=\left(\text{DIAG}_{\text{d}},\text{Mul}_{\text{DIAG}_{\text{d}}},\text{Id}_{\text{DIAG}_{\text{d}}}\right)$.

• $\underline{\text{MAT}_{\text{d},\text{d}}}:=\left(\text{MAT}_{\text{d},\text{d}},\text{Mul}_{\text{MAT}_{\text{d},\text{d}}},\text{Id}_{\text{MAT}_{\text{d},\text{d}}}\right)$.

The former two are commutative. The natural transformations $\text{Det}_{\text{d}}$, $\text{Inc}'_{\text{d}}$, and $\text{Inc}''_{\text{d}}$ (and thus $\text{Inc}_{\text{d}}$) are accordingly the underlying maps of internal monoid morphisms

• $\underline{\text{Det}_{\text{d}}}\ \colon\ \underline{\text{MAT}_{\text{d},\text{d}}}\to\underline{\text{EL}}$.

• $\underline{\text{Inc}'_{\text{d}}}\ \colon\ \underline{\text{EL}}\to\underline{\text{DIAG}_{\text{d}}}$.

• $\underline{\text{Inc}''_{\text{d}}}\ \colon\ \underline{\text{DIAG}_{\text{d}}}\to\underline{\text{MAT}_{\text{d},\text{d}}}$.

(End Proposition 1.)

"Proof" 1.: Similarly direct. $\Box$

Proposition 2.: The copresheaves $\mathbb{1}$, $\text{EL}$, $\text{DIAG}_{\text{d}}$, and $\text{MAT}_{\text{d},\text{d}}$ are corepresented by the crings

• $\mathbb{Z}$

• $\mathbb{Z}\left[a\right]$

• $\mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}$

• $\mathbb{Z}\left[a_{\text{i}_{1},\ \text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}$

respectively, namely via the isomorphisms

• $\left(f\ \colon\ \mathbb{Z}\to R\right)\ \mapsto\ \left(*\ \colon\ \mathbb{1}\left(R\right)\right)$

• $\left(f\ \colon\ \mathbb{Z}\left[a\right]\to R\right)\ \mapsto\ \left(f\left(a\right)\ \colon\ \text{EL}\left(R\right)\right)$

• $\left(f\ \colon\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}\to R\right)\ \mapsto\ \left(\left[\begin{cases}f\left(a_{\text{i}_{0}}\right)\text{ if }\text{i}_{1}=\text{i}_{0} \\ 0\text{ if }\text{i}_{1}\neq\text{i}_{0}\end{cases}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\ \colon\ \text{DIAG}_{\text{d},\text{d}}\left(R\right)\right)$

• $\left(f\ \colon\ \mathbb{Z}\left[a_{\text{i}_{1},\ \text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\to R\right)\ \mapsto\ \left(\left[f\left(a_{\text{i}_{1},\ \text{i}_{0}}\right)\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\ \colon\ \text{MAT}_{\text{d},\text{d}}\left(R\right)\right)$

natural in cring $R$.

"Proof" 2.: Once more by inspection (if you're bored enough to do it). $\Box$

Proposition 3.: Given a commutative monoid $\Lambda$ internal to the coCartesian monoidal structure of $\textbf{CRing}$ and a pair of internal monoid morphisms $$\underline{\partial_{0}},\ \underline{\partial_{1}}\ \colon\ \underline{\text{MAT}_{\text{d},\text{d}}}\ \to\ \Lambda$$ with respective underlying natural transformations $\partial_{0}$ and $\partial_{1}$, we have that $$\partial_{0}\circ\text{Inc}_{\text{d}}\ =\ \partial_{1}\circ\text{Inc}_{\text{d}}\ \implies\ \partial_{0}\circ\text{Inc}''_{\text{d}}=\partial_{1}\circ\text{Inc}''_{\text{d}}\text{.}$$

(I.e., if $\underline{\partial_{0}}$ and $\partial_{1}$ agree after pulling back to $\underline{\text{EL}}$, then they already agreed upon pulling back to $\underline{\text{DIAG}_{\text{d}}}$.)

Proof 3: Let $$\mathfrak{D}_{\text{d}}\ :=\ \left[\begin{cases}a_{\text{i}_{0}}\text{ if }\text{i}_{1}=\text{i}_{0} \\ 0\text{ if }\text{i}_{1}\neq\text{i}_{0}\end{cases}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\ \colon\ \text{DIAG}_{\text{d}}\left(\mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}\right)$$ be the $\text{d}\times\text{d}$ diagonal matrix coclassified by the identity morphism of $\mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}$ via the isomorphisms in Proposition 2, and moreover let $$\left(\mathfrak{D}_{\text{d},\text{i}'}\ =\ \left[\begin{cases}a_{\text{i}_{0}}\text{ if }\text{i}_{1}=\text{i}_{0}=\text{i}' \\ 1\text{ if }\text{i}_{1}=\text{i}_{0}\neq\text{i}' \\ 0\text{ if }\text{i}_{1}\neq\text{i}_{0}\end{cases}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\ \colon\ \text{DIAG}_{\text{d}}\left(\mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}\right)\right)_{\text{i}'\ \colon\ \text{d}}$$ $$\left(\widetilde{\mathfrak{D}_{\text{d},\text{i}'}}\ =\ \text{Inc}'_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(a_{\text{i}'}\right)\ \colon\ \text{DIAG}_{\text{d}}\left(\mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}\right)\right)_{\text{i}'\ \colon\ \text{d}}\text{.}$$ Given $\partial_{0}$, $\partial_{1}$ as in the statement of the proposition, we have by the Yoneda lemma that $$\partial_{0}\circ\text{Inc}''_{\text{d}}\ =\ \partial_{1}\circ\text{Inc}''_{\text{d}}\ \iff\ \partial_{0,\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\circ\text{Inc}''_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(\mathfrak{D}_{\text{d}}\right)\ =\ \partial_{1,\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\circ\text{Inc}''_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(\mathfrak{D}_{\text{d}}\right)\text{.}$$ Indeed, remark that each $\mathfrak{D}_{\text{d},\text{i}'}$ is similar to the corresponding $\widetilde{\mathfrak{D}_{\text{d},\text{i}'}}$ (namely via the invertible $\text{d}\times\text{d}$ matrix that transposes the basis vectors indexed by $0$ and by $\text{i}'$) and thus by the commutativity of $\Lambda$ and the multiplicativity (and thus similarity-invariance) of the components of $\text{Inc}'_{\text{d}}$, $\text{Inc}''_{\text{d}}$, $\partial_{0}$, and $\partial_{1}$ that \begin{align*} \partial_{0,\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\circ\text{Inc}''_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(\mathfrak{D}_{\text{d}}\right)\ &=\ \partial_{0,\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\circ\text{Inc}''_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(\mathfrak{D}_{\text{d},\text{d}-1}\cdots\mathfrak{D}_{\text{d},0}\right)\\ &=\ \prod_{\text{i}'\ \colon\ \text{d}}\text{}^{\Lambda}\ \partial_{0,\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\circ\text{Inc}''_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(\mathfrak{D}_{\text{d},\text{i}'}\right)\\ &=\ \prod_{\text{i}'\ \colon\ \text{d}}\text{}^{\Lambda}\ \partial_{0,\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\circ\text{Inc}''_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(\widetilde{\mathfrak{D}_{\text{d},\text{i}'}}\right)\\ &=\ \prod_{\text{i}'\ \colon\ \text{d}}\text{}^{\Lambda}\ \partial_{0,\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\circ\text{Inc}_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(a_{\text{i}'}\right)\\ &=\ \partial_{0,\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\circ\text{Inc}_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(\prod_{\text{i}'\ \colon\ \text{d}} a_{\text{i}'}\right)\\ &=\ \partial_{1,\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\circ\text{Inc}_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(\prod_{\text{i}'\ \colon\ \text{d}} a_{\text{i}'}\right)\\ &=\ \dots\left(\substack{\text{symmetric} \\ \text{manipulations}}\right)\\ &=\ \partial_{0,\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\circ\text{Inc}''_{\text{d},\ \mathbb{Z}\left[a_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}}\left(\mathfrak{D}_{\text{d}}\right) \end{align*} (with $\prod^{\Lambda}$ the commutative finitary multiplication of $\Lambda$), as claimed. $\Box$

Henceforth denote by $\left[-,-\right]$ the Hom (co)monoid functors defined/discussed in Fact B.1 below (we neglect to annotate the subscript, as context will make the typing evident).

Proposition 4.: $\underline{\text{Inc}_{\text{d}}}$ corepresents an isomorphism on the fully faithful subcategory of commutative monoids internal to the coCartesian monoidal structure of $\textbf{CRing}$.

I.e., given a commutative monoid $\Lambda$ internal to the coCartesian monoidal structure of $\textbf{CRing}$, the map $$\left[\underline{\text{Inc}_{\text{d}}},\ \text{id}_{\Lambda}\right]\ \colon\ \left[\underline{\text{MAT}_{\text{d},\text{d}}},\ \Lambda\right]\ \to\ \left[\underline{\text{EL}},\ \Lambda\right]$$ is an isomorphism of commutative ($\textbf{Set}$) monoids.

Proof 4: It suffices to verify the claim at the level of elements (i.e., of Hom sets). At the very least, the underlying map of $\left[\underline{\text{Inc}_{\text{d}}}, \text{id}_{\Lambda}\right]$ is surjective on elements, seeing as $$\left[\underline{\text{Inc}_{\text{d}}},\ \text{id}_{\Lambda}\right]\circ\left[\underline{\text{Det}_{\text{d}}},\ \text{id}_{\Lambda}\right]\ =\ \left[\text{id}_{\Lambda},\ \text{id}_{\underline{\text{EL}}}\right]\ =\ \text{id}_{\left[\underline{\text{EL}},\ \Lambda\right]}\text{.}$$

It remains to show injectivity, i.e. that given a pair of internal monoid morphisms $$\underline{\partial_{0}},\ \underline{\partial_{1}}\ \colon\ \underline{\text{MAT}_{\text{d},\text{d}}}\ \to\ \Lambda$$ with respective underlying natural transformations $\partial_{0}$ and $\partial_{1}$ that satisfy $\partial_{0}\circ\text{Inc}_{\text{d}} = \partial_{1}\circ\text{Inc}_{\text{d}}$, we have that $\partial_{0}=\partial_{1}$.

Henceforth fix $\partial_{0}$ and $\partial_{1}$ as above. It follows from Proposition 3 that $\partial_{0}\circ\text{Inc}''_{\text{d}} = \partial_{1}\circ\text{Inc}''_{\text{d}}$. The remainder of this proof will be dedicated to deducing the subclaim that $\partial_{0}=\partial_{1}$ given this last equality. (I.e., that if $\underline{\partial_{0}}$ and $\underline{\partial_{1}}$ agree after pulling back to $\underline{\text{DIAG}_{\text{d}}}$, then they already agreed on $\underline{\text{MAT}_{\text{d},\text{d}}}$.)

To that end, let $$\mathfrak{A}_{\text{d}}\ :=\ \left[a_{\text{i},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\ \colon\ \text{MAT}_{\text{d},\times{d}}\left(\mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\right)$$ be the $\text{d}\times\text{d}$ matrix coclassified by the identity of $\mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}$ via the isomorphisms in Proposition 2.

As in the proof of Proposition 3, it suffices by the Yoneda lemma to verify that $$\partial_{0,\ \mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}}\left(\mathfrak{A}_{\text{d}}\right)\ =\ \partial_{1,\ \mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}}\left(\mathfrak{A}_{\text{d}}\right)\text{.}$$ If $\mathfrak{A}_{\text{d}}$ were in the image of $\text{Inc}_{\text{d},\ \mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}}$ (i.e., if it were diagonal), then the subclaim would follow immediately from the Yoneda lemma. Of course, $\mathfrak{A}_{\text{d}}$ is not diagonal, nor even diagonalizable over $\mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}$. Nevertheless, it is diagonalizable in a monic extension of $\mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}$, and we will leverage this fact to deduce the result.

Given a cring $R$ and a $\text{d}\times\text{d}$ matrix $A$ over $R$, the coefficients of the (monomial, degree $\text{d}$) polynomial $\text{det}\left(x-A\right)$ in the indeterminate $x$, so also the discriminant of that polynomial, are expressible as polynomial functions with integer coefficients in the entries of $A$. In particular, there is for each $\text{i}\ \colon\ \text{d}$ a natural transformation $$\text{Coeff}_{\text{d},\text{i}}\ \colon\ \text{MAT}_{\text{d},\text{d}}\to\text{EL}$$ that at the component of $R$ sends the $\text{d}\times\text{d}$ matrix $A$ over $R$ to the coefficient of $x^{\text{i}}$ in $\text{det}\left(x-A\right)$, as well as a natural transformation $$\text{Disc}_{\text{d}}\ \colon\ \text{MAT}_{\text{d},\text{d}}\to\text{EL}$$ that at the component of $R$ sends the $\text{d}\times\text{d}$ matrix $A$ over $R$ to the discriminant of $\text{det}\left(\lambda-A\right)$. What's more, these $\left(\text{Coeff}_{\text{d},\text{i}}\right)_{\text{i}\ \colon\ \text{n}}$ and $\text{Disc}_{\text{d}}$ are by the Yoneda lemma classified by elements $\left(\mathfrak{c}_{\text{d},\text{i}}\right)_{\text{i}\ \colon\ \text{d}}$ and $\mathfrak{d}_{\text{d}}$ of $\mathbb{Z}\left[a_{i_{1},\ i_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}$; these classifying elements are, of course, none other than their respective evaluations on $\mathfrak{A}_{\text{d}}$.

Consider the evident composition of ring morphisms $$\mathbb{Z}\left[a_{\text{i}_{1},\ \text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\ \overset{\varepsilon_{\text{d}}}{\longrightarrow}\ \text{Frac}\left(\mathbb{Z}\left[a_{\text{i}_{1},\ \text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\right)\ \overset{\gamma_{\text{d}}}{\longrightarrow}\ \text{Frac}\left(\mathbb{Z}\left[a_{\text{i}_{1},\ \text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\right)\left[x_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}\ /\ \left(\mathfrak{c}_{\text{d},\text{i}}-\overbrace{\underbrace{\sum_{\text{i}_{0}\ \colon\ \text{d}}\cdots\sum_{\text{i}_{\text{d}-\text{i}-1}\ \colon\ \text{i}_{\text{d}-\text{i}-2}}}_{\text{d}-\text{i}\text{ sums}}\prod_{\text{j}\ \colon\ \text{d}-\text{i}} x_{\text{i}_{\text{j}}}}^{\text{d}-\text{i}^{\text{th}}\text{ elem' symm' poly' in }x\text{s}}\right)_{\text{i}\ \colon\ \text{d}}\text{.}$$ The last ring is readily seen to (essentially by construction) corepresent pairs of $\text{d}\times\text{d}$ matrices $A$ over $R$ equipped with an (ordered) factorization into monomials of $\text{det}\left(x-A\right)$. In particular, it is nontrivial (as there is at least one matrix over at least one nontrivial ring admitting such a datum), so we can choose a(n arbitrary but henceforth fixed) field $\mathfrak{K}_{\text{d}}$ and ring morphism $$\text{Frac}\left(\mathbb{Z}\left[a_{\text{i}_{1},\ \text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\right)\left[x_{\text{i}}\right]_{\text{i}\ \colon\ \text{d}}\ /\ \left(\mathfrak{c}_{\text{d},\text{i}}-\sum_{\text{i}_{0}\ \colon\ \text{d}}\cdots\sum_{\text{i}_{\text{d}-\text{i}-1}\ \colon\ \text{i}_{\text{d}-\text{i}-2}}\prod_{\text{j}\ \colon\ \text{d}-\text{i}} x_{\text{i}_{\text{j}}}\right)_{\text{i}\ \colon\ \text{d}}\ \overset{\kappa_{\text{d}}}{\longrightarrow}\ \mathfrak{K}_{\text{d}}\text{.}$$ By similar logic (specifically that there is at least one $\text{d}\times\text{d}$ matrix $A$ over at least one nonzero ring $R$ with the property that $\text{det}\left(x-A\right)$ has a nonzero discriminant), $\mathfrak{d}_{\text{d}}$ is nonzero and is thus inverted in $\text{Frac}\left(\mathbb{Z}\left[a_{\text{i}_{1},\ \text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}\right)$. So $\kappa_{\text{d}}\circ \gamma_{\text{d}}\circ\varepsilon_{\text{d}}$ is an injective (being the composition of the canonical inclusion $\varepsilon_{\text{d}}$ of a domain into its fraction field with a morphism $\kappa_{\text{d}}\circ\gamma_{\text{d}}$ of fields) morphism from $\mathbb{Z}\left[a_{\text{i}_{1},\ \text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}$ into $\mathfrak{K}_{\text{d}}$ that inverts $\mathfrak{d}_{\text{d}}$.

I.e., the matrix $\text{MAT}_{\text{d},\text{d}}\left(\kappa_{\text{d}}\circ \gamma_{\text{d}}\circ\varepsilon_{\text{d}}\right)\left(\mathfrak{A}_{\text{d}}\right)$ over $\mathfrak{K}_{\text{d}}$ coclassified by $\kappa_{\text{d}}\circ \gamma_{\text{d}}\circ\varepsilon_{\text{d}}$ has the property that $\text{det}\left(x-\text{MAT}_{\text{d},\text{d}}\left(\kappa_{\text{d}}\circ \gamma_{\text{d}}\circ\varepsilon_{\text{d}}\right)\left(\mathfrak{A}_{\text{d}}\right)\right)$ splits as a product of distinct linear monomials (having nonzero discriminant) over $\mathfrak{K}_{\text{d}}$. So by Fact A.2., $\text{MAT}_{\text{d},\text{d}}\left(\kappa_{\text{d}}\circ \gamma_{\text{d}}\circ\varepsilon_{\text{d}}\right)\left(\mathfrak{A}_{\text{d}}\right)$ diagonalizes over $\mathfrak{K}$, being in particular similar to $$\text{Inc}''_{\text{d}}\left(\underbrace{\left[\begin{cases}x_{\text{i}_{0}}\text{ if }\text{i}_{0}=\text{i}_{1} \\ 0\text{ if }\text{i}_{0}\neq\text{i}_{1}\end{cases}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}}_{:=\ \mathfrak{X}_{\text{d}}\ \colon\ \text{DIAG}_{\text{d}}\left(\mathfrak{K}_{\text{d}}\right)}\right)\ \colon\ \text{MAT}_{\text{d},\text{d}}\left(\mathfrak{K}_{\text{d}}\right)\text{.}$$ It follows from the naturality and multiplicativity (whence componentwise similarity-invariance) of $\partial_{0}$ and $\partial_{1}$ that \begin{align*} \text{EL}\left(\kappa_{\text{d}}\circ \gamma_{\text{d}}\circ\varepsilon_{\text{d}}\right)\circ\partial_{0,\ \mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}}\left(\mathfrak{A}_{\text{d}}\right)\ &=\ \partial_{0,\ \mathfrak{K}_{\text{d}}}\circ\text{MAT}_{\text{d},\text{d}}\left(\kappa_{\text{d}}\circ \gamma_{\text{d}}\circ\varepsilon_{\text{d}}\right)\left(\mathfrak{A}_{\text{d}}\right)\\ &=\ \partial_{0,\ \mathfrak{K}_{\text{d}}}\circ \text{Inc}''_{\text{d},\ \mathfrak{K}_{\text{d}}}\left(\mathfrak{X}_{\text{d}}\right)\\ &=\ \partial_{0,\ \mathfrak{K}_{\text{d}}}\circ \text{Inc}''_{\text{d},\ \mathfrak{K}_{\text{d}}}\left(\mathfrak{X}_{\text{d}}\right)\text{,}\\ &=\ \dots\left(\substack{\text{symmetric} \\ \text{manipulations}}\right)\\ &=\ \text{EL}\left(\kappa_{\text{d}}\circ \gamma_{\text{d}}\circ\varepsilon_{\text{d}}\right)\circ\partial_{1,\ \mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}}\left(\mathfrak{A}_{\text{d}}\right) \end{align*} and thus by the monicity of $\kappa_{\text{d}}\circ \gamma_{\text{d}}\circ\varepsilon_{\text{d}}$ and hence of $\text{EL}\left(\kappa_{\text{d}}\circ \gamma_{\text{d}}\circ\varepsilon_{\text{d}}\right)$ that $$\partial_{0,\ \mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}}\left(\mathfrak{A}_{\text{d}}\right)\ =\ \partial_{1,\ \mathbb{Z}\left[a_{\text{i}_{1},\text{i}_{0}}\right]_{\text{i}_{0},\ \text{i}_{1}\ \colon\ \text{d}}}\left(\mathfrak{A}_{\text{d}}\right)\text{,}$$ precisely the subclaimed equality. $\Box$

Remark 5.: See here.

Result 6.: There is an isomorphism $$\left[\underline{\text{MAT}_{\text{d},\text{d}}},\ \underline{\text{EL}}\right]\ \overset{\simeq}{\longrightarrow}\ \left(\mathbb{N},+,0\right)$$ sending $\text{Det}_{\text{d}}$ to $1\ \colon\ \mathbb{N}$.

Proof 6.: By Proposition 2, Fact B.0, and Fact B.2, $\left[\underline{\text{EL}}, \underline{\text{EL}}\right]$ is isomorphic to $\left[\underline{\mathbb{Z}\left[a\right]}, \underline{\mathbb{Z}\left[a\right]}\right]$ with $\underline{\mathbb{Z}\left[a\right]}$ the commutative comonoid internal to the coCartesian monoidal category $\left(\textbf{Cring},\otimes,\mathbb{Z}\right)$ whose underlying object is $\mathbb{Z}\left[a\right]$, whose counit is the cring morphism $$\mathbb{Z}\left[a\right]\overset{a\ \mapsto\ 1}{\longrightarrow}\mathbb{Z}\text{,}$$ and whose binary comultiplication is the map $$\mathbb{Z}\left[a\right]\overset{a\ \mapsto\ bc}{\longrightarrow}\mathbb{Z}\left[b,c\right]\simeq \mathbb{Z}\left[a\right]\otimes \mathbb{Z}\left[a\right]\text{.}$$ By definition, $\left[\underline{\mathbb{Z}\left[a\right]},\ \underline{\mathbb{Z}\left[a\right]}\right]$ is (isomorphic to) the commutative monoid of univariate polynomials $f\ \colon\ \mathbb{Z}\left[a\right]$ satisfying $f\left(1\right)=1$ and $f\left(bc\right)=f\left(b\right)f\left(c\right)$ identically in the indeterminates $b$, $c$, with the monoid operation polynomial multiplication (and unit the constant polynomial at $1$). Thus we obtain by virtue of Fact C.0 an isomorphism of commutative monoids $$\left[\underline{\text{EL}},\ \underline{\text{EL}}\right]\ \overset{\simeq}{\longrightarrow}\ \left[\underline{\mathbb{Z}\left[a\right]},\ \underline{\mathbb{Z}\left[a\right]}\right]\ \overset{\simeq}{\longrightarrow}\ \left(\mathbb{N},+,0\right)$$ mapping the identity of $\underline{\text{EL}}$ to $1\ \colon\ \mathbb{N}$.

Proposition 4. gives us in turn an isomorphism of commutative monoids $$\left[\underline{\text{Inc}_{\text{d}}},\ \underline{\text{id}_{\text{EL}}}\right]\ \colon\ \left[\underline{\text{MAT}_{\text{d},\text{d}}},\ \underline{\text{EL}}\right]\ \overset{\simeq}{\longrightarrow}\ \left[\underline{\text{EL}},\ \underline{\text{EL}}\right]$$ mapping $\underline{\text{Det}_{\text{d}}}$ to $\underline{\text{Det}_{\text{d}}}\circ \underline{\text{Inc}_{\text{d}}} = \text{id}_{\underline{\text{EL}}}$.

Forming the composite $$\left[\underline{\text{MAT}_{\text{d},\text{d}}},\ \underline{\text{EL}}\right]\ \overset{\simeq}{\longrightarrow}\ \left[\underline{\text{EL}},\ \underline{\text{EL}}\right]\ \overset{\simeq}{\longrightarrow}\ \left[\underline{\mathbb{Z}\left[a\right]},\ \underline{\mathbb{Z}\left[a\right]}\right]\ \overset{\simeq}{\longrightarrow}\ \left(\mathbb{N},+,0\right)$$ of the above maps gives the claimed commutative $\textbf{Set}$ monoid isomorphisms. $\blacksquare$

Upshot 7.: $\text{Det}_{\text{d}}$ is the unique generator of the ($\textbf{Set}$) commutative monoid of internal monoid morphisms $\underline{\text{MAT}_{\text{d},\text{d}}}\to\underline{\text{EL}}$, and in particular can be retroactively canonically defined as such.

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A. Some relevant facts from (very elementary) linear algebra, which I painstakingly detail so as to make clear exactly which properties of $\text{det}$ we're presupposing:

See here.

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B. Some relevant facts from category theory:

See here.

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C. A relevant polynomial functional equation:

See here.

Answer 3

This is the appendix to my other answer, separated from the main post due to the character limit.

As before, fix a positive natural $\text{d}$ (the $\text{d}=0$ case is trivial but sadly also degenerate); as before, we abbreviate "$\in$" as "$\colon$" and identify the notation of a natural $\text{n}$ with the set $\left\{0,\dots,\text{n}-1\right\}$.

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A stray remark that didn't fit in the original answer:

Remark 5.: Denote by $\text{MAT}_{\text{d},\text{d}}^{\times}$ the functor sending crings to invertible $\text{d}\times\text{d}$ matrices therein; endow it with the evident internal group structure $\underline{\text{MAT}_{\text{d},\text{d}}^{\times}}$ wrt the Cartesian monoidal structure on $\textbf{CRing}\to\textbf{Set}$ so that $\text{EL}$ and $\text{MAT}_{\text{d},\text{d}}$ become internal $\underline{\text{MAT}_{\text{d},\text{d}}^{\times}}$-modules via the trivial and conjugation actions respectively. The proof of Proposition 4 establish also the tangential claim that $\underline{\text{MAT}_{\text{d},\text{d}}^{\times}}$-module morphisms from the latter to any trivial action are determined by the restriction along $\text{Inc}''_{\text{d}}$ of their underlying natural transformations. (I.e., the morphism of internal modules with underlying map $\text{Inc}''_{\text{d}}$ is "epic with respect to trivial actions".)

In particular, it then follows from the fundamental theorem of symmetric polynomials that the conjugation-invariant natural transformations $\text{MAT}_{\text{d},\text{d}}\to\text{EL}$ are characterized as the evaluations of polynomials with integer coefficients in the coefficients of the characteristic polynomial of a given $\text{d}\times\text{d}$ matrix. (The details are left as an exercise.)

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A. Some relevant facts from (very elementary) linear algebra, which I painstakingly detail so as to make clear exactly which properties of $\text{det}$ we're presupposing:

Fact A.0.: If $\left(A_{\text{j}}\right)_{\text{j}\ \colon\ \text{m}}$ is a sequence of $\text{d}\times\text{d}$ matrices with entries in a commutative ring $R$, then $$\text{det}\left(A_{\text{m}-1}\cdots A_{\text{0}}\right)\ =\ \prod_{\text{j}\ \colon\ \text{m}}\text{det}\left(A_{\text{j}}\right)\text{.}$$ In particular (i.e., in the $\text{m}=0$ case), $\text{det}\left(\text{Id}_{\text{d},R}\right)=1$ where $\text{Id}_{\text{d},R}$ is the $\text{d}\times\text{d}$ identity matrix over $R$.

Proof A.0.: I guess this depends on how you initially define $\text{det}$, but it is in any case fairly straightforward. $\Box$

Fact A.1.: If $A$ is a $\text{d}\times\text{d}$ matrix over some field $K$ and $\text{det}\left(A\right)=0$, then there exists some nonzero $v\ \colon K^{\text{d}}$ such that $Av=0$.

Proof A.1.: Suppose to the contrary that there is no nonzero $v\ \colon k^{\text{d}}$ such that $Av=0$; i.e., that $A$ is injective. Then we have by the dimension theory of vector spaces that $A$ is invertible, from which it follows by Fact A.0 that $\text{det}\left(A^{-1}\right)\text{det}\left(A\right)=1$, which implies in turn that $\text{det}\left(A\right)\neq 0$, contradiction. $\Box$

Fact A.2.: If $A$ is a $\text{d}\times\text{d}$ matrix over some field $K$ and $p_{A}\left(x\right):=\text{det}\left(x-A\right)$ splits as a product of $\text{d}$ distinct linear monomials over $K$ in the indeterminate $x$ (with $x$ shorthand for $x\cdot\text{Id}_{\text{d},K}$ in the former expression), then $A$ diagonalizes (i.e., can be conjugated by an invertible matrix to obtain a diagonal matrix) over $K$.

Proof A.2.: Let $\left(x_{\text{i}}\right)_{\text{i}\ \colon\ \text{d}}$ be the $\text{d}$ distinct roots of $p_{A}\left(x\right)$, and for each $x_{\text{i}}$ choose (as made possible by Fact A.1) nonzero vectors $\left(v_{\text{i}}\ \colon\ K^{\text{d}}\right)_{\text{i}\ \colon\ \text{d}}$ such that $\left(\left(x_{\text{i}}-A\right)v_{\text{i}}=0\right)_{\text{i}\ \colon\ \text{d}}$.

We claim that the tuple of vectors $\left(v_{\text{i}}\right)_{\text{i}\ \colon\ \text{d}}$ is linearly independent. Indeed, observe that for any tuple $\left(q_{\text{i}}\right)_{\text{i}\ \colon\ \text{d}}$ of coefficients satisfying $$\sum_{\text{i}\ \colon\ \text{d}} q_{\text{i}}v_{\text{i}}\ =\ 0\text{,}$$ we have for each $\text{i}\ \colon \text{d}$ that $$0\ =\ \left(x_{\text{d}-1}-A\right)\cdots\left(x_{\text{i}+1}-A\right)\cdot \left(x_{\text{i}-1}-A\right)\cdots\left(x_{0}-A\right)\sum_{i\ \colon\ \text{d}} q_{\text{i}}v_{\text{i}}\ =\ \left(\prod_{\substack{\text{i}\ \colon\ \text{d} \\ \text{i}'\ \neq\ \text{i}}} x_{\text{i}'}-x_{\text{i}}\right) q_{\text{i}}v_{\text{i}}$$ from which it follows, using the fact that $v_{\text{i}}\neq 0$ and $\left(\prod_{\substack{\text{i}'\ \colon\ \text{d} \\ \text{i}'\ \neq\ \text{i}}} x_{\text{i}'}-x_{\text{i}}\right)\neq 0$, that $q_{\text{i}}=0$ for all $\text{i}\ \colon\ \text{d}$.

By the dimension theory of vector spaces $\left(v_{\text{i}}\right)_{\text{i}\ \colon\ \text{d}}$ is then moreover a basis of $K^{\text{d}}$, each member of which manifestly satisfies $Av_{\text{i}} = q_{\text{i}}v_{\text{i}}$. So the conjugation of $A$ by the change-of-basis-to-$\left(v_{\text{i}}\right)_{\text{i}\ \colon\ \text{d}}$ matrix yields the diagonal matrix with diagonal entries $\left(\lambda_{\text{i}}\right)_{\text{i}\ \colon\ \text{d}}$, and $A$ is in particular diagonalizable as claimed. $\Box$

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B. Some relevant facts from category theory:

In this section, fix a category $\mathcal{C}$ with all finitary coproducts (the coproduct of $\mathcal{C}$ written as $+$, its initial object $0$), and denote by

• $\textbf{Set}$ the category of sets.

• $\underline{\mathcal{C}} := \left(\mathcal{C},+,0\right)$ the coCartesian monoidal structure on $\mathcal{C}$.

• $\underline{\widehat{\mathcal{C}}}:=\left(\mathcal{C}\to\textbf{Set},\times,1\right)$ the Cartesian monoidal structure on the copresheaf category of $\mathcal{C}$.

Denote in turn by

• $\textbf{CMon}$ the category of commutative monoids (in $\textbf{Set}$).

• $\textbf{coMon}_{\underline{\mathcal{C}}}$ the category of comonoids internal to $\underline{\mathcal{C}}$.

• $\textbf{coCMon}_{\underline{\mathcal{C}}}$ the category of commutative comonoids internal to $\underline{\mathcal{C}}$.

• $\textbf{Mon}_{\underline{\widehat{\mathcal{C}}}}$ the category of monoids internal to $\underline{\widehat{\mathcal{C}}}$.

• $\textbf{CMon}_{\underline{\widehat{\mathcal{C}}}}$ the category of commutative monoids internal to $\underline{\widehat{\mathcal{C}}}$.

Fact B.0.: The contravariant Yoneda embedding $\text{co}よ_{\mathcal{C}}\ \colon\ \mathcal{C}^{\text{op}}\ \to\ \left(\mathcal{C}^{\text{op}}\to\textbf{Set}\right)$ renders the (dual of the) coCartesian monoidal structure of $\underline{\mathcal{C}}$ compatible with the monoidal structure of $\underline{\widehat{\mathcal{C}}}$. In particular, it induces evident fully faithful embeddings

• $\textbf{Mon}_{\text{co}よ_{\mathcal{C}}}\ \colon\ \textbf{coMon}_{\underline{\mathcal{C}}}^{\text{op}}\ \to\ \textbf{Mon}_{\underline{\widehat{\mathcal{C}}}}$

• $\textbf{CMon}_{\text{co}よ_{\mathcal{C}}}\ \colon\ \textbf{coCMon}_{\underline{\mathcal{C}}}^{\text{op}}\ \to\ \textbf{CMon}_{\underline{\widehat{\mathcal{C}}}}$

with respective essential images the fully faithful subcategory of monoids internal to $\underline{\widehat{\mathcal{C}}}$ with corepresentable underlying object and commutative monoids internal to $\underline{\widehat{\mathcal{C}}}$ with corepresentable underlying object.

"Proof" B.0.: There are many subclaims here, but they all follow via straightforward inspection in light of the Yoneda lemma and that Hom preserves limits, so in particular finitary products, (contravariantly) in its first argument. $\Box$

Fact/Definition B.1.: If $\underline{W}=\left(W,\mu_{W},\iota_{W}\right)$ is a commutative comonoid internal to $\underline{\mathcal{C}}$, then for any other (not necessarily commutative) comonoid $\underline{W'}=\left(W',\mu_{W'},\iota_{W'}\right)$ internal to $\underline{\mathcal{C}}$, $\text{Hom}_{\textbf{coMon}_{\underline{\mathcal{C}}}}\left(\underline{W},\underline{W'}\right)$ inherits a commutative monoid structure "copointwise" from $\underline{W}$.

In particular, the unit of this commutative monoid is the composition of $\underline{W}$'s counit map $\iota_{W}\colon W\to 0$ and the initial map $0\to W'$.

Its commutative binary multiplication is given by sending pairs $f_{0},\ f_{1},\ W\to W'$ of underlying maps of internal comonoid morphisms to the composition of $\underline{W}$'s comultiplication map $\mu_{W}\colon W\to W+W$ with the map $\left\langle f_{0},f_{1}\right\rangle\colon W+W\to W$ induced from $f_{0}$ and $f_{1}$ via the universal property of the binary coproduct.

This commutative monoid structure on $\text{Hom}_{\textbf{coMon}_{\underline{\mathcal{C}}}}\left(\underline{W},\underline{W'}\right)$ moreover upgrades via pre-/post-composition to a bifunctor

• $\left[-,-\right]_{\textbf{coMon}_{\underline{\mathcal{C}}}}\ \colon\ \textbf{coCMon}_{\underline{\mathcal{C}}}^{\text{op}}\times \textbf{coMon}_{\underline{\mathcal{C}}}\ \to\ \textbf{CMon}$.

Dually and perhaps more familiarly, an analogous process produces a bifunctor

• $\left[-,-\right]_{\textbf{Mon}_{\underline{\widehat{\mathcal{C}}}}}\ \colon\ \textbf{Mon}_{\underline{\widehat{\mathcal{C}}}}^{\text{op}}\times \textbf{CMon}_{\underline{\widehat{\mathcal{C}}}}\ \to\ \textbf{CMon}$

that agrees at the level of elements with the underlying Hom. (Note that the commutative monoid must now be in the codomain argument.)

"Justification" B.1.: The proof is a diagram.

In fact, we can prove the result almost purely abstractly; we will handle the comonoid case, with the proof of the (hopefully more familiar) monoid case formally dual:

The relevant claim(s) can be expressed entirely in terms of (families) of diagrams involving (universal properties of) finitary coproducts and composition relations, so is reflected by the contravariant Yoneda embedding, and thus true iff the formally dual claims re internal monoids hold in copresheaf categories. But finitary products (note the contravariance!) and compositions are pointwise in presheaf categories, so the general claim is true iff it's true for monoids internal to $\textbf{Set}$, where it's very familiar indeed.

Note that both reductions (from comonoids in $\underline{\mathcal{C}}$ and from monoids in $\underline{\widehat{\mathcal{C}}}$) ultimately end in directly proving facts about monoids in $\textbf{Set}$. Note moreover that the claim concerning $\underline{\widehat{\mathcal{C}}}$ holds for all Cartesian monoidal categories; indeed, we nowhere in the argument use that it's specfically a copresheaf category. $\Box$

Fact B.2.: The obvious diagram relating $\textbf{Mon}_{\text{co}よ_{\mathcal{C}}}$, $\textbf{CMon}_{\text{co}よ_{\mathcal{C}}}$, $\left[-,-\right]_{\textbf{coMon}_{\underline{\mathcal{C}}}}$, and $\left[-,-\right]_{\textbf{Mon}_{\underline{\widehat{\mathcal{C}}}}}$ commutes. I.e., $$\left[-,-\right]_{\textbf{coMon}_{\underline{\mathcal{C}}}}\ \simeq\ \left[-,-\right]_{\textbf{Mon}_{\underline{\widehat{\mathcal{C}}}}}\ \circ\ \left(\textbf{Mon}_{\text{co}よ_{\mathcal{C}}},\ \textbf{CMon}_{\text{co}よ_{\mathcal{C}}}\right)\ \circ\ \varsigma$$ with $\varsigma\ \colon\ \textbf{coCMon}_{\underline{\mathcal{C}}}\times \textbf{coMon}_{\underline{\mathcal{C}}}\ \to\ \textbf{coMon}_{\underline{\mathcal{C}}}\times \textbf{coCMon}_{\underline{\mathcal{C}}}$ the "swap entries" functor.

"Proof" B.2.: This is once again a direct corollary of the Yoneda lemma. (And in particular the fully faithfulness of the Yoneda embedding.) $\Box$

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C. A relevant polynomial functional equation:

(I'd rather write it out here than in the place it's actually used.)

Fact C.0.: The space of univariate polynomials $p\ \colon \mathbb{Z}\left[a\right]$ satisfying $p\left(1\right)=1$ and $p\left(bc\right)=p\left(b\right)p\left(c\right)$ identically in the indeterminates $b$, $c$ is $\left(a^{\text{n}}\right)_{\text{n}\ \colon\ \mathbb{N}}$.

Proof C.0.: First, all elements of $\left(a^{\text{n}}\right)_{\text{n}\ \colon\ \mathbb{N}}$ clearly satisfy the condition. Moreover, the space of polynomials in question is clearly closed under (polynomial) multiplication and, if the quotient is a polynomial, division as well.

Suppose that $p$ satisfies the condition; clearly $p\neq 0$, so that there exist unique $\text{n}\ \colon \mathbb{N}$, nonzero $q\ \colon \mathbb{Z}$ and polynomial $p'\ \colon \mathbb{Z}\left[a\right]$ such that $$p\left(a\right)\ =\ \left(p'\left(a\right)+q\right)a^{\text{n}}$$ and thus by the above that $$p'\left(bc\right)+q\ =\ p'\left(b\right)p'\left(c\right)+q p'\left(b\right)+q p'\left(c\right)+q^{2}$$ identically in the indeterminates $b$, $c$. Comparing coefficients then forces $q p'\left(b\right)=0$ identically in $b$, from which it follows (via the nonzeroness of $q$) that $p'\left(a\right)=0$ identically in $a$ and then (via the evaluation condition) that $q = 1$.

So the list of solutions above is exhaustive. $\Box$