Here are my attempts to prove the following lemma:
(a) Let $f:A\to A'$ be an isomorphism. Let $g:A'\to A$ be its inverse. Consider $$\alpha_{A,A'}: \mathscr A(A,A')\to \mathscr B(J(A),J(A'))$$ This is a bijective map (but it seems that this is irrelevant here). We have $\alpha_{A,A'}(f)=J(f):J(A)\to J(A')$. We need to show that it is an isomorphism.
Consider $$\alpha_{A,A}: \mathscr A(A,A)\to \mathscr B(J(A),J(A))$$
We have $1_A\mapsto J(1_A)=1_{J(A)}$. On the other hand, $g\circ f\mapsto J(g\circ f)=J(g)\circ J(f)$. But $1_A=g\circ f$, and so their images must be equal, i.e. $J(g)\circ J(f)=1_{J(A)}$.Similarly, $J(f)\circ J(g)=1_{J(A')}$. Thus $J(f)$ is an isomorphism.
Addition I missed the other direction when I was writing this, but I think the other direction is proved in (b) below.
Is the above correct? I don't see where I used fullness of faithfulness – are these needed? (Except the "Addition" part.)
For the below, I just wanted to make sure that my proofs are correct.
(b) Let $g:J(A)\to J(A')$ be an isomorphism. The map $\alpha_{A,A'}$ from above is bijection, so there is a unique $f:A\to A'$ in $\mathscr A$ that maps to $g$. It remains to show that $f$ is an isomorphism.
Let $g':J(A')\to J(A)$ be the inverse of $g$. Call its unique preimage under $\alpha_{A',A}$ $f'$. Let's verify that $f'$ is an inverse of $f$.
Consider $f'\circ f:A\to A$. Consider $\alpha_{A,A}:\mathscr A(A,A)\to \mathscr B(J(A),J(A))$. Note $$f'\circ f\mapsto J(f')\circ J(f)=g'\circ g=1_{J(A)}\\ 1_A\mapsto 1_{J(A)}$$ By injectivity, $f'\circ f=1_A$. Similarly, $f\circ f'=1_{A'}$. Thus $f$ is an isomorphism.
(c) If $f:A\to A'$ is an isomorphism, then by (a) $\alpha_{A,A'}(f)=J(f):J(A)\to J(A')$ is an isomorphism. Conversely, if $g:J(A)\to J(A')$ is an isomorphism, then by (b) there is a unique isomorphism $f:A\to A'$. (It seems the condition that the image of $f$ is $g$ is not needed.)
Best Answer
Your proofs all look correct, and you are also correct that you don't need to use fullness or faithfulness to prove that a functor preserves isomorphisms. All functors preserve isomorphisms, exactly by the proof that you gave.
As you've essentially noticed in (b), you need the functor to be fully faithful to prove it reflects isomorphisms, since otherwise you might not be able to lift the inverse or show that the lift of the inverse is an inverse to the original morphism.