I read in [wikipedia][1] that: One can show that a functor $F : C → D$ yields an equivalence of categories if and only if $F$ is full, faithful and essentially surjective.

I'm trying to prove this but I have an issue: Let $F:C\to D$ and $G:D\to C$ be functors and we have natural isomorphisms $\epsilon: FG\to 1_D$ and $\eta: 1_C\to GF$. I need to show that $F$ is full, faithful and essentially surjective.

Essentially surjective is easy: Take $P\in D$, then $P\simeq F(G(P))$ because of $\epsilon$ and we are done.

Faithful is also not hard: Take $f,g: X\to Y$ such that $F(f)=F(g)$, then $GF(f)=GF(g)$ using the commutative diagrams (its hard to draw it here) of $\eta$ and that it is a natural isomorphism we find $f=g$.

But proving $F$ is full is problematic. It all boils down to this map $$ F(X)\xrightarrow{F\eta_Z}FGF(X)\xrightarrow{\epsilon_{F(Z)}}F(X) $$ if this map is identity (as is usually the case for counits and units) then we are done because by commutative diagram of $\epsilon$, for a morphism $\alpha: F(X)\to F(Y)$ we find $$ \alpha = \epsilon_{F(Y)}\: FG(\alpha) \: \epsilon_{F(X)}^{-1} $$ and because $\epsilon_{F(Z)}\circ F\eta_Z=\mathrm{id}$ (and a similar relation in the other direction) we are done since $\alpha$ becomes $F(f)$ for some $f$.

But in the definiton of equivalence of categories we do not demand any special relation between $\epsilon$ and $\eta$ just that they are natural isomorphisms. I'm not completely sure, but I don't think one can prove that $F(X)\xrightarrow{F\eta_Z}FGF(X)\xrightarrow{\epsilon_{F(Z)}}F(X)=\mathrm{id}$ using what is available in the definition. To me there doesn't seem any reason why this is identity.

So can this last formula be proven? Or is there another proof for fullness? [1]: https://en.wikipedia.org/wiki/Equivalence_of_categories


1 Answer 1


Yes, your map is not an identity in general, but you can always get around that problem by replacing one of the isomorphisms of your equivalence in such a way that it becomes an adjoint equivalence (see here).

You don't need to do that, however. Take a morphism $g : FX → FY$. There is only one choice for a morphism $f : X → Y$ such that $g = Ff$: by the naturality of $η$, $f$ must equal $η^{-1} ∘ Gg ∘ η$. So let $f$ be given by the previous formula, from which it follows that $GFf = Gg$. But $G$ is an equivalence, and you've already proven that equivalences are faithful, so $g = Ff$.

  • 12
    $\begingroup$ The step $f = \eta^{-1} \circ Gg \circ \eta \Rightarrow GFf = Gg$ is explained in more detail in this answer. $\endgroup$
    – ComFreek
    Commented Nov 19, 2018 at 16:18

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