Functors preserve isomorphisms

Since $f$ and $g$ are mutually inverse, $g \circ f = 1_A$ and $f \circ g = 1_B$.

Apply $F$ to the first equation. Functors preserve composition, so $F(g \circ f) = F(g) \circ F(f)$; functors preserve identities, so $F(1_A) = 1_{F(A)}$. Combining these with $g \circ f = 1_A$ gives $F(g) \circ F(f) = 1_{F(A)}$.

Apply $F$ to the second equation in the same way: $F(f \circ g) = F(f) \circ F(g)$ and $F(1_B) = 1_{F(B)}$, so from $f \circ g = 1_B$ we get $F(f) \circ F(g) = 1_{F(B)}$.

The two displayed equations say exactly that $F(g)$ is a two-sided inverse of $F(f)$. A morphism with a two-sided inverse is an isomorphism, so $F(f) : F(A) \to F(B)$ is an isomorphism with inverse $F(g)$, as claimed.