The background:

The naive notion of isomorphism of categories is that: a functor $F:\mathscr{C}\rightarrow\mathscr{D}$ is an isomorphism if there exists $F^{-1}:\mathscr{D}\rightarrow\mathscr{C}$ such that $F^{-1}\circ F$ and $F\circ F^{-1}$ are equal to identity map.

I'm starting my studies about category theory from the sight of representations theory. The book that I'm following (Introduction to Representations Theory - Etingof et al) states that this naive notion is not useful and show the following example:

Let $\mathscr{C}_1$ the most simple category. $Ob(\mathscr{C}_1)=\{X\}$ and $\textrm{Hom}(X,X)=\{id_X\}$. Also let $\mathscr{C}_2$ the category that have the objects $X$ and $Y$, and the following four morphisms: $1_x,1_Y,a: X\rightarrow Y,b:Y\rightarrow X$. So we must have $a\circ b=1_Y$ and $b\circ a=1_X$. It easy to check that for any category $\mathscr{D}$ there is a natural bijection between the collections of isomorphism classes of functors $\mathscr{C}_1\rightarrow \mathscr{D}$ and $\mathscr{C}_2\rightarrow \mathscr{D}$.

This show us that for "practical purposes" the categories $\mathscr{C}_1$ and $\mathscr{C}_2$ are the same but using the naive notion they will be not isomorphic (they don't have the same number of objects).

My question:

My doubt maybe is quite simple and I would like if the answer could be detailed for a beginner on this subject. Why is it easy to check that for any category $\mathscr{D}$ there is a natural bijection between the collections of isomorphism classes of functors $\mathscr{C}_1\rightarrow \mathscr{D}$ and $\mathscr{C}_2\rightarrow \mathscr{D}$? How to check this step by step?

  • $\begingroup$ Useful: For a category $C$, if we keep only one object in every isomorphism class, we get a skeleton of $C$. Two categories are equivalent iff their skeletons are isomorphic. (Now the two objects in $\mathscr C_2$ are isomorphic, and neither has nontrivial endomorphism, so $\mathscr C_2$ is equivalent to $\mathscr C_1$.) $\endgroup$
    – Berci
    Mar 28 '16 at 0:08
  • 1
    $\begingroup$ Your textbook has to resort to "equality of isomorphism classes of functors" to justify equivalence of categories to a beginner? Wow! Good luck with the rest of the book. If your purpose is to study category theory, I suggest you get a proper introductory textbook on CT. Awodey, or "the joy of cats" (free online) come to mind. MacLane's CWM is also excellent of course. $\endgroup$
    – magma
    Mar 29 '16 at 11:39

Functors $\mathscr{C}_1\rightarrow \mathscr{D}$ are completely determined by what object $X$ gets sent to. Two such functors $F,G$ are isomorphic iff the objects $F(X)$ and $G(X)$ are isomorphic, indeed the data of an isomorphism $F(X)\cong G(X)$ is all you need to define a natural isomorphism from $F$ to $G$. Hence isomorphism classes of functors $\mathscr{C}_1\rightarrow \mathscr{D}$ are indexed by isomorphism classes of objects in $\mathscr{D}$.

On the other hand, a functor $F:\mathscr{C}_2\rightarrow \mathscr{D}$ is given by a choice of two isomorphic objects $F(X)$ and $F(Y)$ and a choice of an isomorphism between them. This a priori seems like it is keeping track of more information, and it is, however we are looking at isomorphism classes of functors not the functors themselves. The claim now is that two such functors $F, G:\mathscr{C}_2\rightarrow \mathscr{D}$ are isomorphic iff $F(X)$ is isomorphic to $G(X)$.

The "only if" part I think should be clear, so to show the "if" part we want to take an isomorphism $\phi$ (of objects) from $F(X)$ to $G(X)$ and construct a natural isomorphism between $F$ and $G$. Well all that amounts to is constructing an isomorphism from $F(Y)$ to $G(Y)$, which makes the one commutative diagram you can possibly write down commute (I'll leave it to you to actually write it down). Anyway $G(a)\circ \phi \circ F(b)$ should be exactly the isomorphism needed. Hence isomorphism classes of functors $\mathscr{C}_2\rightarrow \mathscr{D}$ are also indexed by isomorphism classes of objects in $\mathscr{D}$.


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