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$\newcommand{\A}{\mathscr{A}}\newcommand{\B}{\mathscr{B}}\newcommand{\C}{\mathscr{C}}$ This is reminiscent of this question I asked ages ago on projective limits of dynamical systems. Note that I have not studied limits in category theory yet and that this was an isolated example.

Take the following exercise from Tom Leinster's introduction to the theory:

Let $F:\A\times\B\to\C$ be a functor. Show that for each $A\in\A$ there is a functor $F^A:\B\to\C$ defined on objects $B\in\B$ by $F^A(b):=F(A,B)$ and on maps $g$ in $\B$ by $F^A(g):=F(1_A,g)$.

To me, someone who comes from backgrounds of analysis, topology and generally subjects not involving much abstract structural work, giving the definition as Leinster has above is already sufficient proof the function exists, after one has observed the function (functor, I should say) is well-defined and satisfies the axioms of a functor. Is the student just supposed to verify axiom satisfaction or is there more going on here?

Relating to the post that I linked, I had great trouble interpreting the question (see $2i)$ and $2ii)$ there) because the question left me with the same strange feeling. To briefly restate the issue, it was to show that: if $K$ is the projective limit of topological dynamical systems, for any different topological dynamical system $L$ there was a unique homomorphism $\tau:L\to K$ satisfying $\pi_i\tau=\sigma_i$, where the $\sigma_i$ were given mappings and $\pi_i$ is the projection from $K$ to its $i$th coordinate. To me, then and now, the question is oddly trivial - since $\pi_i\tau=\sigma_i$ for all $i$ if and only if $\tau=(\sigma_i)_{i\in I}$, uniqueness and existence is immediate, obvious, and I ask what was there for me to actually prove - I refuse to believe the exercise was intended to be that trivial, considering most exercises require more than a sentence of work.

The answer given there was helpful but I did not understand it properly, and I still don't since I am not well versed in category theory yet. It seemed to say that my interpretation is correct but that the exercise is not trivial/ridiculous as I claim.

Tldr: Could someone clarify why the general question:

Show there is a function $F$ with [some specified properties that completely characterise $F$].

Is not immediately and trivially answered by:

Let $F$ have said fully characterising properties and (very quickly!) check it satisfies other contextual axioms on $F$. $\blacksquare$

I think this issue is important to resolve for my continued study of category theory.

I am aware that not all functional equations are satisfied but I don't think that is relevant to this type of issue.

FShrike
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    If someone says "check that $(G,*)$ is a group" you then check that it meets the definition. This isn't any different. That doesn't mean it's always going to be a mind-blowing deep exercise. – Randall May 04 '22 at 18:16
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    Functor $\neq$ function. – Thomas Andrews May 04 '22 at 18:22
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    Here, in the first question, the $F^A$ is defined, but you don’t have proof that it satisfies the properties of a functor. So here, the distinction between functor and function really matters. – Thomas Andrews May 04 '22 at 18:27
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    The conclusion is not that trivial after all. Consider the similarly worded, but wrong "Let $f\colon [0,1]\times[1,2]\to\Bbb R$ be a Lebesgue-measurable function. Show that for each $a\in [0,1]$, there is a Lebesgue measurable function $f^a\colon [1,2]\to\Bbb R$ defined for all $b\in[1,2]$ by $f^a(b)=f(a,b)$" – Hagen von Eitzen May 04 '22 at 18:52
  • @HagenvonEitzen That is quite interesting to me, having studied measure theory, and I've never come across it. Do you have a nice counterexample? – FShrike May 04 '22 at 18:59

2 Answers2

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You have missed out the stated context of the question (1.2.25 in the text).

Sometimes we meet functors whose domain is a product $\mathcal{A} \times \mathcal{B}$ of categories. Here you will show that such a functor can be regarded as an interlocking pair of families of functors, one defined on $\mathcal{A}$ and the other defined on $\mathcal{B}$. (This is very like the situation for bilinear and linear maps.)

The question is a three-part question designed to get you to show something specific; the author could have explained this viewpoint in the text, but they chose instead to set it as a guided exercise. It's not unusual for such a guided exercise to be laid out as a sequence of several trivial steps, to help the reader who is expected to not yet be comfortable with the definitions.

Patrick Stevens
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First, a functor is not a function.

If the categories are all “small,” you can think of a functor as a specific kind of function, but it still has to satisfy some properties, not just be defined on objects and maps.

This is more like defining a function completely, but being asked to prove it is a continuous function. Just defining a function doesn’t ensure it is continuous, and just defining $F^A$ on objects and maps doesn’t mean you’ve defined a functor.

Specifically, you need to show that $F^A(1_B)=1_{F(A,B)}$ and $F^A(g_1\circ g_2)=F^A(g_1)\circ F^A(g_2),$ when $g_1,g_2$ are compatible maps in $\mathscr B.$

There are other times when we define a function which requires a choice. Usually, when we have an equivalence relation $\sim$ on $X$ and we want a function $X/\sim\to Y.$ We often define this by essentially finding $f_1:X\to Y,$ and then we have to show the corresponding $f$ is “well-defined.”

In these cases, we are really trying to show that the value of $f(x)$ is unique of any $x\in X/\sim.$ That is, different representatives of $x$ in $X$ don’t have different values for $f_1.$

This is often first encountered when defining modular arithmetic, or, more generally, quotient groups and rings.

Some people might encounter this first when defining the integers in terms of the natural numbers, or the rationals in terms of the integers.

Thomas Andrews
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