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Homea functor is an equivalence iff it is fully faithful and essentially surjective
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a functor is an equivalence iff it is fully faithful and essentially surjective
Proposition 1.
A functor is an equivalence iff it is full, faithful and essentially surjective.
Proof.
$(\Rightarrow)$. Let functor $F\colon C\to D$ be an equivalence. So there is a functor $G\colon D\rightarrow C$ such that $FG\approxeq 1_{D}$ and $GF\approxeq 1_{C}$, where $\approxeq$ means naturally isomorphic. For $FG\approxeq 1_{D}$, this means that there are natural transformations
$\displaystyle\alpha$  $\displaystyle\colon FG\Rightarrow 1_{D}$  
$\displaystyle\alpha^{*}$  $\displaystyle\colon 1_{D}\Rightarrow FG$ 
such that for any $\overline{x}\in\mathcal{OB}{(}D)$, $\alpha_{{\overline{x}}}\,\alpha^{*}_{{\overline{x}}}=1_{{\overline{x}}}$ and $\alpha^{*}_{{\overline{x}}}\,\alpha_{{\overline{x}}}=1_{{FG(\overline{x})}}$, meaning that $\overline{x}$ and $FG(\overline{x})$ are isomorphic (as objects of $D$). Similarly, for $GF\approxeq 1_{C}$, there are natural transformations
$\displaystyle\beta$  $\displaystyle\colon GF\Rightarrow 1_{C}$  
$\displaystyle\beta^{*}$  $\displaystyle\colon 1_{C}\Rightarrow GF$ 
such that for any $x\in\mathcal{OB}{(}C)$, $\beta_{x}\,\beta^{*}_{x}=1_{x}$ and $\beta^{*}_{x}\,\beta_{x}=1_{{GF(x)}}$, meaning that $x$ and $GF(x)$ are isomorphic (as objects of $C$).
For any $\overline{x}\in\mathcal{OB}{(}D)$, let $x=G(\overline{x})$. To show that $F(x)\cong\overline{x}$ is the same as showing $FG(\overline{x})\cong\overline{x}$. By the previous discussion, this is clear. Therefore, $F$ is essentially surjective.
We now show that $F$ is faithful. Let $f\in\hom(x,y)$ for $x,y\in\mathcal{OB}{(}C)$. Then by the above, there is a commutative diagram
$\xymatrix@C=1.5cm{{x}\ar[r]^{{f}}\ar[d]^{{\beta^{*}_{x}}}&{y}\ar[d]^{{\beta^{*% }_{y}}}\\ {GF(x)}\ar[r]_{{GF(f)}}&{GF(y)}}$ 
Thus $\beta^{*}_{y}\,f=GF(f)\,\beta^{*}_{x}$; applying $\beta_{y}$ to both sides we have $f=\beta_{y}\,GF(f)\,\beta^{*}_{x}$. Thus $f$ is determined by $GF(f)$ so that $GF$ must be faithful. Applying the same argument to a morphism $g\in\hom(\overline{x},\overline{y})$ for $\overline{x},\overline{y}\in\mathcal{OB}{(}D)$ and using $FG$ in place of $GF$ shows that $G$ is faithful as well.
Next we show that $F$ is full. Suppose $g\in\hom(F(x),F(y))$ for $x,y\in\mathcal{OB}{(}C)$. Let $f=\beta_{y}\,G(g)\,\beta^{*}_{x}$. Then the following diagram commutes for either choice of horizontal arrow on the bottom: the top arrow by definition of $\beta$ and $\beta^{*}$; the bottom arrow by definition of $f$.
$\xymatrix@C=1.5cm{x\ar[r]^{f}\ar[d]_{{\beta^{*}_{x}}}&y\ar[d]^{{\beta^{*}_{y}}% }\\ GF(x)\ar@<.5ex>[r]^{{GF(f)}}\ar@<.5ex>[r]_{{G(g)}}&GF(y)}$ 
It follows that $G(g)\,\beta^{*}_{x}=\beta^{*}_{y}\,f=GF(f)\,\beta^{*}_{x}$; composing on the right with $\beta_{x}$ gives $G(g)=GF(f)$. Finally, since $G$ is faithful, we have $g=F(f)$ and $F$ is full.
$(\Leftarrow)$. Now, we assume that $F\colon C\to D$ is essentially surjective, full and faithful. We want to show that $F$ is an equivalence.
Since $F$ is fully faithful, for any $x,y\in\mathcal{OB}{(}C)$, the homsets $\hom(x,y)$ and $\hom(F(x),F(y))$ are bijective. In particular, since $\hom(x,x)\cong\hom(F(x),F(x))$, $1_{x}$ gets bijectively mapped to $1_{{F(x)}}$.
Next, let $a,b\in\mathcal{OB}{(}D)$, we can find $x,y\in\mathcal{OB}{(}C)$ such that $F(x)\cong a$ (via a bijective morphism, say $\alpha\colon F(x)\to a$) and $F(y)\cong b$ (via bijection $\beta\colon F(y)\to b$). So, given $g\colon a\to b$, we have $\beta^{{1}}g\alpha\colon F(x)\to F(y)$. Similarly, for $f\colon F(x)\to F(y)$, we have $\beta f\alpha^{{1}}\colon a\to b$. Therefore, $\hom(a,b)\cong\hom(F(x),F(y))\cong\hom(x,y)$. In particular, for any $a\in\mathcal{OB}{(}D)$ with $x\in\mathcal{OB}{(}C)$ such that $F(x)\cong a$, since $\hom(a,a)\cong\hom(x,x)$, $1_{a}$ gets mapped bijectively to $1_{x}$.
If, for any $a\in\mathcal{OB}{(}D)$, there are $x_{1},x_{2}\in\mathcal{OB}{(}C)$ such that $F(x_{1})\cong a\cong F(x_{2})$, then $\hom(x_{1},x_{2})\cong\hom(a,a)\cong\hom(x_{2},x_{1})$. Suppose $1_{a}$ gets mapped (bijectively) to $s\colon x_{1}\to x_{2}$ and to $t\colon x_{2}\to x_{1}$. Because $\hom(a,a)\cong\hom(a,a)\times\hom(a,a)\cong\hom(x_{1},x_{2})\times\hom(x_{2},x% _{1})$, $1_{a}$ gets mapped to $st\colon x_{1}\to x_{1}$. But $1_{a}$ is also mapped bijectively to $1_{{x_{1}}}\colon x_{1}\to x_{1}$, this shows that $st=1_{{x_{1}}}$. Similarly, $ts=1_{{x_{2}}}$. So, we have $x_{1}\cong x_{2}$.
Define $[x]=\{y\mid y\cong x\}$ and pick a representative $x_{0}$ with a bijection $x\to x_{0}$. Any $a\in\mathcal{OB}{(}D)$ can be mapped bijective into members of $[x]$ can be mapped bijectively to $x_{0}$. So, for any $a\in[a]\subseteq\mathcal{OB}{(}D)$, we have a welldefined $G(a):=x_{0}$.
For any morphism $g\colon a\to b$, $a,b\in\mathcal{OB}{(}D)$, pick representatives $a_{0}\in[a]$ and $b_{0}\in[b]$, then there is a unique morphism $g^{*}\colon a_{0}\to b_{0}$ given by $\xymatrix@C=0.5cm{{a_{0}}\ar[r]&{a}\ar[r]^{{g}}&{b}\ar[r]&{b_{0}}}$. Since $\hom(a,b)\cong\hom(a_{0},b_{0})\cong\hom(x_{0},y_{0})$, $g^{*}$ in turns gets mapped to a unique morphism in $\hom(x_{0},y_{0})$. We call this $G(g)$.
Now, we are ready to define the functor $G\colon D\to C$. For an object $a$, we define $G(a)$ to be the unique $x_{0}\in\mathcal{OB}{(}C)$ discussed two paragraphs ago. For a morphism $g\colon a\to b$, we define $G(g)$ to be the unique morphism $f\colon G(a)\to G(b)$ that is found in the previous paragraph. In order for this to be a functor, we need to check the following:

$G(1_{a})=1_{{G(a)}}$, and

$G(g_{1}g_{2})=G(g_{1})G(g_{2})$.
These can be easily verified, as follows: $1_{a}$ is mapped to
$\xymatrix@C=1cm{{a_{0}}\ar[r]&{a}\ar[r]^{{1_{a}}}&{a}\ar[r]&{a_{0}}={a_{0}}\ar% [r]^{}{1_{{a_{0}}}}&{a_{0}}},$ 
and $1_{{a_{0}}}$ is mapped to $1_{{x_{0}}}=1_{{G(a)}}$. The second property can be verified similarly.
Finally, we want to show that $FG\approxeq 1_{D}$ and $GF\approxeq 1_{C}$. This amounts to showing that for each $x\in\mathcal{OB}{(}C)$, $GF(x)\cong x$, and for each $a\in\mathcal{OB}{(}D)$, $FG(a)\cong a$. For the first part, $x$ gets mapped to $F(x)$ which is in turn mapped to $GF(x)=x_{0}$, where $F(x_{0})\cong F(x)$, which means that $x_{0}\cong x$, or that $GF(x)\cong x$. For the second part, $a$ is mapped first to $G(a)=x_{0}$, where $F(x_{0})\cong a$. Then $x_{0}$ is in turn mapped to $F(x_{0})$. Thus, $FG(a)=F(x_{0})\cong a$. ∎
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