# Is this functor not fully faithful and essentially surjective or are these categories equivalent?

I know that there are proofs that a fully faithful and essentially surjective functor realises an equivalence of categories. There are hints how to prove that e.g. here and here.

However, I made up an example that seems like a counter-example to that statement and at the moment I just can not spot my (probably trivial) mistake, so please help me to point it out.

The example is as follows: Consider the following diagram:

and take $A$ and $B$ to be objects of category $\mathcal{C}$. Let them be sets where $A$ has 3 elements and $B$ has 2 elements and let $\phi$ send these 3 elements to only one element of $B$ and let $\psi$ send the 2 elements of $B$ to one of the elements of $A$.

Now let $C$ be the object of category $\mathcal{C}_2$, let it also be a set with one element and let the functor $F$ send the objects and arrows as indicated below the diagram.

Is there already an error in this construction? If not, one can show that $F$ is fully faithful and essentially surjective:

A functor is said to be faithful/full iff for every $A,A'\in\mathcal{C}$, the induced map $F:\text{Mor}(A,A')\rightarrow\text{Mor}(F(A),F(A'))$ is injective/surjective.

This is the case because there is always exactly one morphism from any object $A\in\mathcal{C}$ to any other object $A'\in\mathcal{C}$ which can then be mapped to the identity.

A functor is said to be essentially surjective iff for every object $C$ in $\mathcal{C}_2$, there is an object $A\in\mathcal{C}$ such that $F(A)=C$.

This is true too by definition of $F$. However,

A functor $F:\mathcal{C}\rightarrow\mathcal{C}_2$ is said to realise an equivalence iff there is a functor $G:\mathcal{C}_2\rightarrow\mathcal{C}$ such that $F\circ G\cong 1_{\mathcal{C}}$ and $G\circ F\cong 1_{\mathcal{C}_2}$.

Now I do not see which functor could be constructed to show the equivalence because if $G$ sends $C$ to $G(C)=A$, then $G(F(B))=A$ and $A$ is not isomorphic to $B$. The same holds if $G(C)=B=G(F(A))$.

What is it that I do not take into consideration properly? Thank you.

• Hint: There is only one morphisms from $C \to C$, but there are multiple morphisms from $A to A$ (such as $\psi \circ \phi$ which is not the identity). – user45878 Nov 7 '17 at 19:06
• Ouh...one could consider the composition of $\psi\circ\phi$ as additional maps from $A$ to $A$. Ouh man, thank you, that is it! – exchange Nov 7 '17 at 19:09
• @user45878 Haha, I knew it was trivial, sometimes the forest is hidden behind the trees. Okay, if you want, you can post it as answer or do you think I should delete the question altogether? – exchange Nov 7 '17 at 19:10
• @exchange It's a well-posed question, so I'd suggest you write a quick answer for completeness. – Hanno Nov 7 '17 at 19:22

For example, $\psi\circ\phi$ sends the 3 elements of the set $A$ to a single element of $A$ and is therefore an additional element of $\text{Mor}(A,A)$ which means that the functor $F$ is not faithful anymore because it sends $F(\psi\circ\phi)$ to $F(\psi)\circ F(\phi)=id_{C}$.