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I have found a proof here that shows that two notions of categorical equivalence are equivalent to each other. But I don't understand the line of reasoning, in particular the sections below which I have colored red:

4.1 Theorem. Let $\mathcal{C}$ and $\mathcal{D}$ be categories and let $F : \mathcal{C} \rightarrow \mathcal{D}$ be a covariant functor. Then $F$ is an equivalence iff $F$ is full, faithful, and dense.

Proof. ($\Rightarrow$) Suppose that $F$ is an equivalence and that $G$, $\eta$, and $\mu$ are as above. $\color{red}{\textrm{Then the map $f \mapsto GF(f)$ is a bijection (isomorphism) from $Mor_\mathcal{C}(GF(C), GF(C'))$.}}$ Thus the map $f \mapsto F(f)$ is injective (monic) from $\text{Mor}_\mathcal{C}(C, C')$ to $\text{Mor}_\mathcal{D}(F(C), F(C')$, and the map $G : F(f) \mapsto GF(f)$ from $\text{Mor}_\mathcal{D}(F(C), F(C'))$ to $\text{Mor}_\mathcal{C}(GF(C), GF(C'))$ is surjective. $\color{red}{\textrm{Using the natural isomorphism $\mu$ gives that $f \mapsto F(f)$ is surjective (epic).}}$ Thus $F$ is full and faithful. $\color{red}{\textrm{That it is dense is trivial.}}$

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I assume $\eta: 1_{\mathcal C}\to GF$ and $\mu: 1_{\mathcal D}\to FG$ are the natural isomorphisms.

Then, all arrows $\eta_c:c\to GF(c)$ in $\mathcal C$ and $\mu_d:d\to FG(d)$ in $\mathcal D$ are isomorphisms, i.e. they have unique inverses.

  1. By naturality of $\eta$, for an $f:c\to x$ we have that $GF(f)\circ \eta_c = \eta_x\circ f$, i.e. $$\begin{align} GF(f) &= \eta_x\circ f\circ\eta_c^{-1}\\ f &= \eta_x^{-1}\circ GF(f)\circ\eta_c \end{align}$$ By the way, this map $f\mapsto GF(f)$ actually goes from the homsets $\hom_{\mathcal C}(c,x)\ \ $ to $\hom_{\mathcal C}(GF(c),\,GF(x))$.

  2. So, we get $G\mid_{\hom(F(c),F(x))}:\hom(F(c),F(x))\to\hom(GF(c),GF(x))$ is surjective (mapping $u\mapsto G(u)$ for any $u:F(c)\to F(x)$).
    Basically, we can repeat the same for $\mu$ to obtain that $FG$ is bijective on homsets, so $F$ will be surjective on homsets.

  3. For any $d\in\mathrm{Ob}(\mathcal D)$, we have $F(G(d))\cong d$.

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    RE (2): When repeating the argument on $FG$, we don't establish that $FG$ is surjective, we establish that $GFf \mapsto FGFf$ is surjective. But how do we know that every arrow $f$ in $\mathcal{C}$ has representation $GFf'$ for some $f'$? – 2017-02-17
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    This all is symmetric in $F$ and $G$. By 1., we get that $GF$ restricted to homsets is bijection. Repeating the arguments with $F$ and $G$ exchanged, we get that $FG$ is bijection. For an arrow $f:c\to x$, define your $f'$ as $\eta_x^{-1}\circ f\circ\eta_c$. By the equations written at 1., it will follow that $GF(f')=f$. – 2017-02-18
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    Oh I see. In the article I linked, $\mu : FG \rightarrow 1_\mathcal{D}$. But in your answer it is $\mu : 1_\mathcal{D} \rightarrow FG$. Does this matter? – 2017-02-18