If $ A \stackrel{f} {\longrightarrow} B \stackrel{g} {\longrightarrow} C \stackrel{h} {\longrightarrow} D$ is an exact sequence, prove that $f$ is surjective if and only if $h$ is injective.
Proof ($\longrightarrow$): Suppose $f$ is surjective. Then $Im(f) = B$. Apply the first isomorphism theorem for exact sequences, then $B / Im(f) \cong C$. By exactness $Im(f) = ker(g)$. Then $B/Im(f) = B/Ker(g) \cong Im(g)$. So, $B/Im(f) \cong Im(g)$ and hence that $B/Im(f) \cong Ker(h)$ (since $Im(g) = Ker(h)$). Since $Im(f) = B$, then $B/Im(f) = B/B = \left\{{0}\right\} \cong Ker(h)$.Therefore, $Ker(h) \cong \left\{{0}\right\}$ which implies that $h$ is injective.
For the proof in the reverse direction it looks I do something similar with assuming $h$ injective and then using exactness and possibly cokernels. I'm not sure if the reverse direction of the proof might be a copy of the forward direction with certain modifications.
Is this proof decent? How would I approach the other direction? Other question is whether or not it is legal for me to use the fact in the forward proof that $B/Im(f) \cong Im(g)$. I understand that $Im(g) \cong C$ could only potentially hold if $g$ is surjective. If so, how would I remedy this situtation?
Advice and tips would be greatly appreciated.
Your forward proof looks more complicated than it has to be. We can do it without quoting isomorphism theorems.
Since $f$ is surjective and the sequence is exact, the kernel of $g$ is all of $B$ and so the image of $g$ is $\{0\}$. Since the kernel of $h$ equals the image of $g$, we get that $h$ is injective.
Conversely, if $h$ is injective then its kernel is $\{0\}$ which is the image of $g$. Since $g$ maps all of $B$ to $0$ we get that the kernel of $g$ is $B$ which is the image of $f$. Therefore, $f$ is surjective.