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If $F,G: C \rightarrow D$ are functors between regular categories and $F \Rightarrow G$ is a natural isomorphism, is it true that if $F$ is faithful {resp. full) then $G$ is full? I believe this is true but would like some confirmation.

What about if $F$ is essentially surjective? Is it necessarily true that $G$ is essentially surjective?

Thanks.

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1 Answer 1

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Let $C$ and $D$ be categories (arbitrary), $F,G\colon C\to D$ functors, $\alpha\colon F\to G$ a natural isomorphism. Then:

  • If $F$ is faithful, then $G$ is faithful.

Proof: Let $c_1,c_2\in C$ be objects and $f_1,f_2:c_1\to c_2$ morphisms in $C$. Then, by naturality: $$ G(f_i)\circ \alpha(c_1)=\alpha(c_2)\circ F(f_i),\quad i=1,2. $$ We know that $\alpha$ is an isomorphism, so we have: $$ G(f_i)=\alpha(c_2)\circ F(f_i)\circ(\alpha(c_1))^{-1},\quad i=1,2. $$ Therefore, if $G(f_1)=G(f_2)$, then $F(f_1)=F(f_2)$, and finally $f_1=f_2$.

  • If $F$ is full, then $G$ is full.

Proof: Let $c_1,c_2\in C$ be objects and $g\colon G(c_1)\to G(c_2)$ a morphism in $D$. Let's define the morphism $k\colon F(c_1)\to F(c_2)$ as follows: $$ k=(\alpha(c_2))^{-1}\circ g\circ\alpha(c_1). $$ $F$ is full, therefore there exists a morphism $f\colon c_1\to c_2$ in $C$, such that $F(f)=k$. It's easy to check that $G(f)=g$.

  • If $F$ is essentially surjective, then $G$ is essentially surjective.

Proof: Let $d\in D$ be object, then there exists such $c\in C$, that $d\cong F(c)$. But $F(c)\cong G(c)$ by $\alpha$, thus, by transitivity, $d\cong G(c)$.

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