Identity about a Functor

Question

I'm moving my first steps in CT and suddenly after reading about functors, this question came up in my mind:

Let $F \colon A \to B$ a functor between two categories $A$ and $B$, is it true that for any arrow $f \colon a \to a'$ in $A$,

is commutative?

According to me the definition of functor doesn't involve nor implies this identity. Nevertheless, due to my restricted experience in CT, I cannot find a counterexample.

I hope the clarification will help me build a strong "set" of example/counterexample in this field.

Thanks in advance.

Answer 1

If $F\colon A\longrightarrow B$ is a functor from the category $A$ to the category $B$, then it associates to each object $a$ of $A$ an object $F(a)$ of $B$ and to each arrow $f\colon a\rightarrow a'$ in $A$ an arrow $F(f)\colon F(a)\rightarrow F(a')$ in $B$. This association is made in a way compatible with the identity morphisms in $A$ and with the composition law in $A$.

Thus, given an object $a$ in $A$, $F(a)$ is an object of $B$. Hence, in general, there can be no arrow from $a$ to $F(a)$, simply because the two objects live in two different categories! Therefore, the square you built can not be understood meaningfully in a general situation. Even if $A=B$ (so that $F(a)$ is an object of $A$ for all objects $a$ in $A$), there may still be no sensible way to "connect $a$ to $F(a)$ via the action of $F$" (i.e. to get something as the "arrow" $a\stackrel{F}{\rightarrow} F(a)$ you put in the square). For example, for a prescribed arrow $h\colon a\rightarrow F(a)$ in $A$ (seen as the codomain of $F$), there might be no arrow $g\colon a''\rightarrow a$ in $A$ (seen as the domain of $F$) such that $F(a'')=a$ and $F(g)=h$.

The only way I see to give a meaning to your square, in general, is in the following situation. Suppose you have a functor $F\colon A\longrightarrow A$ (what is called an endofunctor of $A$) and a natural transformation $\tau\colon Id_{A}\Rightarrow F$ from the identity functor of $A$ into $F$ (if you do not know what a natural transformation of functors is, then just check the definition in any book of Category Theory!). Then, for each arrow $f\colon a\rightarrow a'$ in $A$, you do have $F(f)\circ \tau (a)=\tau(a')\circ f$, just by definition of natural transformation.

Hope this helps somehow.

Answer 2

In addition to Marco's answer, let me remark that one might construct such an endofunctor with a natural transformation (on a bigger category) for any functor $F:{\bf A}\to{\bf B}$, based on exactly the commutative squares in your question!

So, take the disjoint union of ${\bf A}$ and ${\bf B}$ and take new basic arrows $j_a:a\to F(a)$, and define all compositions formally so that the commutativities of your squares hold.

If we want to write it, we can say that the new arrows ('heteromorphisms') $a\to b$ are just pairs $\langle \beta,a\rangle$ where $\beta:F(a)\to b$, and then $j_a=\langle 1_{F(a)},a\rangle$, and composition by an $\alpha:a_0\to a$ is defined as $$\langle \beta,a\rangle\circ\alpha :=\langle\ \beta\circ F(\alpha),\,a_0\rangle\,.$$

[Note that this is just the (collage of) the profunctor $F_*:{\bf A}^{op}\times{\bf B}\to{\bf Set}$ defined by $(a,b)\mapsto\hom_{\bf B}(F(a),b)$]