I’d like to ask for checking of my attempt below.
We want to find $g: A_2 \to A_1$ such that $f \circ g = 1_{A_2}$ and $g \circ f = 1_{A_1}$. So define $g: A_2 \to A_1$ to be a morphism such that $F(g \circ f) = 1_{F(A_1)}$.
We first look at $(g \circ f)$. Let $h_1: A_1 \to A_2$ be an arbitrary morphism. First we want to show that $h_1 \circ (g \circ f) = h_1$.
Now:
\begin{align} F(h_1 \circ (g \circ f))
&= F(h_1) \circ F(g \circ f) \ \ \ \text{(since $F$ is covariant)} \\
&= F(h_1) \circ 1_{F(A_1)} \ \ \ \ \ \ \ \text{(by definition of $g$)} \\
&= F(h_1)
\end{align}
Then, since $F$ is injective: $h_1 \circ (g \circ f) = h_1$.Similarly, we can let $h_2: A_2 \to A_1$ be an arbitrary morphism, and show that $(g \circ f) \circ h_2 = h_2$.
The above shows that $g \circ f = 1_{A_1}$.
To carry out the analysis for $(f \circ g)$ in a similar fashion, I’d need the fact that $F(f \circ g) = 1_{F(A_2)}$. I show it as follows:
Since $F(f): F(A_1) \to F(A_2)$ is an isomorphism, there exists $F’: F(A_2) \to F(A_1)$ such that $F(f) \circ F’ = 1_{F(A_2)}$ and $F’ \circ F(f) = 1_{F(A_1)}$.
Now from the definition of $g$, we have:
\begin{align}
& \ \ \ F(g \circ f) = 1_{F(A_1)} = F’ \circ F(f) \\
& \Rightarrow \ \ \ \ \ \ F(g) \circ F(f) = F’ \circ F(f) \\
& \Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F(g) = F’
\end{align}
Then: $$F(f \circ g) = F(f) \circ F(g) = F(f) \circ F’ = 1_{F(A_2)}$$
I have several questions:
$1. $ Where is the assumption that $F$ is surjective used? I’m tempted to think that it enables $g$ to be defined the way it is: since $1_{F(A_1)} \in \text{Hom}(F(A_1),F(A_1))$, and $F$ is surjective, there exists $\varphi \in \text{Hom}(A_1,A_1)$ such that $F(\varphi) = 1_{F(A_1)}$. So $\varphi$ justifies the existence of $(g \circ f)$. I’d appreciate if other instances like that, if any, could be pointed out in my proof.
$2. $ I’m also asked to show that the injectivity of $F$ alone does not suffice. Is the above a good explanation for that?
.
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Best Answer
If $F(f):F(A_1)\to F(A_2)$ is an isomorphism, then this means there is $g':F(A_2)\to F(A_1)$ such that $g'\circ F(f)=1_{F(A_1)}$ and $F(f)\circ g'=1_{F(A_2)}$. Since $F$ is full, there is $g:A_2\to A_1$ with $F(g)=g'$. Now $$F(g\circ f)=F(g)\circ F(f)=g'\circ F(f)=1_{F(A_1)}$$ But also $F(1_{A_1})=1_{F(A_1)}$, so we must have $g\circ f=1_{A_1}$ since $F$ is faithful. A similar argument shows that $f\circ g=1_{A_2}$.