Question about why category of sets is not equivalent to its dual.

Question

I am trying to understand why $Set$ (category of sets) is not equivalent to $Set^{op}$.

To do so we show that

1. Every morphism into the initial object in Set is an isomorphism
2. There is a morphism into an initial object in $Set^{op}$ which is not an isomorphism.

Is my reasoning correct as to why we can conclude the result:

Suppose $F$ is a functor involved in an equivalence between Set and $Set^{op}$. Let $F(\emptyset) = \{*\}$ an initial object in $Set^{op}$ (equivalences preserve colimits). Then $F$ is fully faithful and essentially surjective. Now take a set $A$ such that $A\rightarrow \{*\}$ is not an isomorphism in $Set^{op}$. There should be a set $B$ such that $F(B)$ is isomorphic to $A$. So we get a map $\psi: F(B)\rightarrow \{*\}$. Clearly this cannot be an isomorphism (else the previous map would be one too). So we have a map, $\psi: F(B)\rightarrow F(\emptyset)$ and $F$ full means there is a map $f:B\rightarrow \emptyset $ such that $F(f) = \psi$. But if $f$ is an isomorphism, then so is $F(f) = \psi$, a contradiction.

Is this reasoning correct?

Answer 1

Your proof is correct. There is just one detail that may need some justification. You say:

Now take a set $A$ such that $A \to \{*\}$ is not an isomorphism in $Set^{op}$.

You should probably justify why $A$ and such an arrow exist.

Of course, that is quite simple, take for example $A = \{0,1\}$ and take the function $\{*\} \to \{0,1\}$ sending $*$ to $0$. This is clearly not an isomorphism in $Set$, so this gives us an arrow $\{0,1\} \to \{*\}$ in $Set^{op}$ that is not an isomorphism.

Now, this was just me being nitpicky, but since the entire proof relies on that single fact it may be worth being more explicit about it.