3
$\begingroup$

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 $\ \ \ \log_{|B|}|GB|=\log_{|B|}|B^I|=|I|$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Zhen Lin Jun 25 '14 at 14:36
  • 1
    $\begingroup$ 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 : $2^\omega = 2^1$. $\endgroup$ – Pece Jun 25 '14 at 15:20
  • 1
    $\begingroup$ @ZhenLin: So I think $|B|^{|FA|}=|B^{FA}|=|\mathrm{hom}_{\mathcal{C}}(FA,B)|=|\mathrm{hom}_{\mathcal{D}}(A,GB)|=|GB|^{|A|}$ and then $|FA|=|A|\cdot\log_{|B|}|GB|$. $\endgroup$ – Nikolaj-K Jun 25 '14 at 22:13
  • 1
    $\begingroup$ @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. $\endgroup$ – Pece Jun 26 '14 at 8:23
  • 1
    $\begingroup$ @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 ? $\endgroup$ – Pece Jun 26 '14 at 8:50
0
$\begingroup$

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 $b^{F(a)} \cong G(b)^a$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Nikolaj-K Jun 26 '14 at 7:59
  • 1
    $\begingroup$ I think you're assuming $C=D$? $\endgroup$ – Hurkyl Jun 26 '14 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.