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: enter image description here

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.

  • 2
    $\begingroup$ 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). $\endgroup$ Nov 7, 2017 at 19:06
  • 1
    $\begingroup$ Ouh...one could consider the composition of $\psi\circ\phi$ as additional maps from $A$ to $A$. Ouh man, thank you, that is it! $\endgroup$
    – exchange
    Nov 7, 2017 at 19:09
  • $\begingroup$ @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? $\endgroup$
    – exchange
    Nov 7, 2017 at 19:10
  • 1
    $\begingroup$ @exchange It's a well-posed question, so I'd suggest you write a quick answer for completeness. $\endgroup$
    – Hanno
    Nov 7, 2017 at 19:22

2 Answers 2


Okay, the thing I had not taken into consideration properly was that composition of maps away from an object and back to an object can ultimately be viewed as maps from that object to itself, too.

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 $F(\psi)\circ F(\phi)=id_{C}$.

As a result, there shouldn't be any equivalence and the example turn out to satisfy the theorem that a functor realises an equivalence iff it is fully faithful and essentially surjective.


A delayed answer, but it is perhaps good to set the record straight on this question. In short: yes, $F$ is an equivalence of categories and $A$ is indeed isomorphic to $B$.

In your answer, you seem to suggest that the morphism $\psi \circ \phi$ is different from $\operatorname{id}_A$. However, I don't think it is: the four arrows are probably the only four morphisms the category has and the composition of $\phi$ with $\psi$ should give the identity on $A$.

With this interpretation of the example, the functor $F$ is both essentially surjective and it is fully faithful, precisely because of the reasons you mention.

This is not a contradiction, because the functor $F$ is indeed an equivalence. There are two choices for an inverse $G$, of which one is the one you gave: $G(C) = A$, $G(\operatorname{id}_C) = \operatorname{id}_A$. As you point out, $G(F(B)) = A$. But note that $A$ is isomorphic to $B$ via $\phi: A \to B$ with inverse $\psi: B \to A$! So $G \circ F$ is naturally equivalent to the identity on the category $\mathcal{C}_1$! Also clearly $F \circ G$ is equal to the identity on $\mathcal{C}_2$, so $G$ is an inverse of $F$ and thus $F$ is an equivalence of categories.

On a final note, let me remark that your definition of essentially surjective is not quite correct. A functor is said to be essentially surjective iff for every object $C$ in $\mathcal{C}_2$, there is an object $A \in \mathcal{C}_1$ such that $F(A)$ is isomorphic to $C$! It is important that $F$ does not have to be strictly surjective on objects! For example, the functor $G: \mathcal{C}_2 \to \mathcal{C}_1$ is essentially surjective, even though it is not surjective.


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