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Here is a very silly question:

Adjoint functors satisfy

$$\mathrm{hom}_{\mathcal{C}}(FA,B) \cong \mathrm{hom}_{\mathcal{D}}(A,GB).$$

I consider numbers $a,b$ and read this as

$$b^{\,f(a)}=g(b)^a.$$

If the objects in the categories can be assigned cardinalities, do the functors actually fulfill a relation along those lines?

$\bf Edit$: If e.g. $|B^{FA}|=|B|^{|FA|}$ does make sense, just taking the cardinalities of the hom-sets tells us

$$\frac{|FA|}{|A|}=\log_{|B|}|GB|.$$

E.g. in a category with object being sets, the adjoint functors $FA:=A\times I$ and $GB:=B^I$ have

$\frac{|FA|}{|A|}=\frac{|A\times I|}{|A|}=|I|\ \ \ $ and $\ \ \ \log_{|B|}|GB|=\log_{|B|}|B^I|=|I|$.

Najib Idrissi
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Nikolaj-K
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  • Well, if the categories in question are concrete, we can just use the usual set-theoretic notion of cardinality. But then we can't say very much about what the functors do, even if we know they are adjoint. – Zhen Lin Jun 25 '14 at 14:36
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    Well it depends largely on what you means by "the objects in the categories can be assigned cardinalities". For example take $\mathcal C$ the category of groups and $\mathcal D$ the category of sets. Consider the adjunction $F \dashv U$ where $U \colon \mathcal C \to \mathcal D$ is the forgetful functor and $F$ is the free group functor. As cardinalities of objects, I take cardinalities of the (underlying) sets. Then, for the group $\mathbb Z/2$, and a singleton ${x}$, we certainly do not have : $2^\omega = 2^1$. – Pece Jun 25 '14 at 15:20
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    @ZhenLin: So I think $|B|^{|FA|}=|B^{FA}|=|\mathrm{hom}{\mathcal{C}}(FA,B)|=|\mathrm{hom}{\mathcal{D}}(A,GB)|=|GB|^{|A|}$ and then $|FA|=|A|\cdot\log_{|B|}|GB|$. – Nikolaj-K Jun 25 '14 at 22:13
  • Pece: Yeah, the natural thing is to consider the cardinalities of the hom-sets (and not a priori interpret the hom-sets as numerical exponents). Then if $A={x},\ FA=\mathbb Z$, we have $|\mathrm{hom}{\bf{Grp}}(FA,B)|=1$ because $\mathbb Z$ is initial and $|\mathrm{hom}{\bf{Set}}(A,UB)|=1$ because the singleton is terminal. :) – Nikolaj-K Jun 26 '14 at 07:54
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    @NikolajK But didn't you want to assign cardinalities to objects? Then what does it means for a group $B$ to set $|B|$ as the "cardinality of the hom-set" : hom-set from/to what ? There is something unclear here. – Pece Jun 26 '14 at 08:23
  • @Pece: To objects if they are the images of a functor, yes. I think if I can't do cardinal arithmetic on any side of the adjunction, then I can't do much. But if I can, then I can relate the two by either taking the $|B|$-logarithm or taking the $|A|^\mathrm{th}$ root. – Nikolaj-K Jun 26 '14 at 08:33
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    @NikolajK Ok, but it doesn't clear the ambiguity out. Reconsider my example of the adjunction $F \dashv U$. What cardinality do you assign to $\mathbb Z = F(1)$ ? Because, $|\hom(\mathbb Z, 1)| = 1$ but $|\hom(\mathbb Z,\mathbb Z/2)| = 2$ ($\mathbb Z$ is not initial as you said, you must confuse with $\mathsf{Ring}$). So, what is $|F(1)|$ here ? – Pece Jun 26 '14 at 08:50
  • @Pece: Right, both cardinalities of the homs I wrote down are in fact 2, so the relation is true. And yeah, I can't write down a formula for a number assigned $F(1)$, I only get to $|U\mathbb Z_2|=|\mathrm{hom}_{\bf{Grp}}(F(1),\mathbb Z_2)|$. – Nikolaj-K Jun 26 '14 at 09:15

1 Answers1

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I don't know if we can interpret this as an equation of numbers, but what about the following: For objects $a,b$ of a category the exponential $b^a$ is an object with a natural bijection $\hom(c,b^a) \cong \hom(c \times a,b)$. A category is called cartesian closed if it has all binary products and exponentials for any two objects. Now assume that $C$ is cartesian closed. Assume that $F : C \to C$ is left adjoint to $G$. We want to compare $b^{F(a)}$ with $G(b)^a$. Thus, we want to compare $$\hom(c,b^{F(a)})=\hom(c \times F(a),b)$$ with $$\hom(c,G(b)^a) = \hom(c \times a,G(b)) = \hom(F(c \times a),b)$$ Therefore, a natural isomorphism $b^{F(a)} \cong G(b)^a$ corresponds to a natural isomorphism $c \times F(a) \cong F(c \times a)$. Taking $a$ to be the terminal object $1$ and letting $x=F(1)$, we get that $F(c) \cong c \times x$. It follows $G(b) \cong b^x$. Therefore, the isomorphism reduces to the plausible isomorphism $$b^{a \times x} \cong (b^x)^a.$$ This can be also checked directly. For arbitrary $F$, I don't know how to make sense of $b^{F(a)} \cong G(b)^a$.

Martin Brandenburg
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  • Thanks, I think I figured out some good relations in the comments above. And I'd read it as: if $F$ and $G$ are adjoint, then $G$ has an exponentially big image, as compared to $F$. It's because of the sides the stand in the hom-relation, domain vs. codomain. – Nikolaj-K Jun 26 '14 at 07:59
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    I think you're assuming $C=D$? –  Jun 26 '14 at 08:25