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Let $2=\{a,b\}$ be any set with exactly $2$ elements $a$ and $b$. Define a functor $F: {\bf{Sets}}/2\rightarrow{\bf{Sets}}\times{\bf{Sets}}$ with $F(f:X\rightarrow 2)=(f^{-1}(a),f^{-1}(b))$. Is this an isomorphism of categories?

This is something that I've done: In ${\bf{Sets}}/2$, let $h:X\rightarrow\{a,b\}$ such that $h^{-1}(a)\not=\emptyset$ and $h^{-1}(b)\not=\emptyset$ and let $f:X\rightarrow\{a,b\}$ such that $f^{-1}(a)=X$ and $f^{-1}(b)=\emptyset$. Then there is no arrow between $f$ and $h$ in ${\bf{Sets}}/2$, but there is one arrow between $F(f)=(X,\emptyset)$ and $F(h)=(h^{-1}(a),h^{-1}(b))$ in ${\bf{Sets}}\times{\bf{Sets}}$, so these two categories are not isomorphic.

Is this reasoning ok?

I am a complete novice in the category theory!

Jovana
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2 Answers2

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Your argument is not correct. In fact, $F$ is fully faithful.

Define a functor $G : \mathsf{Set} \times \mathsf{Set} \to \mathsf{Set} / \{a,b\}$ by mapping two sets $A,B$ to the function $G(A,B) : A \coprod B \to \{a,b\}$ which maps the elmenets of $A$ to $a$ and the elements of $B$ to $b$. Here, $A \coprod B= A \times \{1\} \cup B \times \{2\}$ is the standard construction of the disjoint union. Then we have $FG \cong \mathrm{id}$ and $GF \cong \mathrm{id}$. Hence, $F$ is an equivalence of categories.

But $F$ is not an isomorphism. In fact, $F$ is not surjective on objects. If $A,B$ are two sets which are not disjoint, then $(A,B)$ doesn't lie in the image of $F$. (Notice, however, that $F$ lies in the essential image of $F$.)

Here is how I think about this: The notion of "isomorphism of categories" belongs to $0$-category theory $\approx$ set theory. It is much too strong to explain various "structural equalities of categories". A far better behaved notion is that of an "equivalence of categories". Only this belongs to $1$-category theory. If $A,B$ are two sets, then the property that $A,B$ are disjoint cannot be formulated in category theoretic terms, since this property is not invariant under self-equivalences of $\mathsf{Set} \times \mathsf{Set}$. In fact, we can make $A,B$ disjoint by replacing them by $A^* = A \times \{1\}$ and $B^* = B \times \{2\}$. Then $(A,B) \cong (A^*,B^*)$ in $\mathsf{Set} \times \mathsf{Set}$. In category theory, in order to state that two objects $A,B$ are disjoint, we need a larger object $S$ which contains both. Then their intersection is given by the pullback $A \times_S B$, and $A,B$ are called disjoint when this pullback is an initial object. In set theory, one often imagines the universe $V$ to be some kind of universal object which contains all other objects of consideration. Every set then has a unique embedding into $V$. In category theory, however, one really has to remember all the embeddings. In particular, there is a completely different point of view on subsets. For more on this, see SE/704593 and SE/295800. In my opinion category theory is the better language to explain structural mathematics.

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Martin Brandenburg
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  • Another approach that I have tried comes down to the fact that F is not surjective, but I was not sure whether I am allowed to use this principle (on an arbitrary large category) to prove that the mapping is not an isomorphism. So, whenever we talk about the "isomorphism of categories", we can reason like we do in set theory? I was wondering how should someone who has, let's say, read only the first chapter of Awodey's book and then tried to do corresponding exercises, think about this issue. Thank you so much for your answer! – Jovana Jul 14 '14 at 13:56
  • Look up the definition of an isomorphism of categories - you will immediately realize that it induces a bijection on the object classes. – Martin Brandenburg Jul 14 '14 at 14:39
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I'm afraid your argument is not correct: there are morphisms from $f$ to $h$.

Hint: Perhaps $F$ is not an $isomorphism$ of categories, but something slightly weaker?

Daniel Gerigk
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