Rotman's Homological Algebra Ex 9.31 I have a question about ex 9.31 in Rotman's text. It looks like a standard diagram chasing proof but I seem to be missing something.  The problem is below,

For part iii, I think we want to show ker$(\overline{d}) \cong $ ker$(\alpha)$ induced by the map $f$, so we need to check that $f(\text{im} \Delta) = 0$.
But the only thing that I can think of to show that is $g \alpha f (\Delta(a)) = d \Delta(a) = 0$ and since $g$ is injective, $\alpha f(\Delta(a)) = 0$.  Since we don't know anything about the map $\alpha$, I don't see how we can conclude that $f(\text{im} \Delta) = 0$ though.
I tried to use part ii, so we have $g \phi = \overline{d}$ and since $g$ is injective, we can conclude that $\ker{\overline{d}} = \ker(\phi) = \ker(\alpha f)$, I don't think this helps as well.
Any hints would be appreciated, thank you.
 A: As in my first comment, there are missing assumptions required for (iii) to work. As in my second comment, assuming merely $g$ to be injective is not enough to make (iv) work. As a counterexample for the latter: Take all modules except $C$ to be ${\mathbb Z}$, and let $C={\mathbb Z}\oplus{\mathbb Z}$ with $\Delta=d=\alpha=0$, $f={\rm id}$ and $g(c’)=(c’,0)$ for $c’\in C’.$
In sum, to make everything work, the natural additional assumptions are ${\rm im}(\Delta)={\rm ker}(f)$ and that $g$ is an isomorphism. The proof is straightforward or via applying the Snake Lemma to the following diagrams: $$\begin{array}{ccccccccc}&&A&\rightarrow&B&\rightarrow&B’&\rightarrow&0 \\ &&\downarrow&&\downarrow&&\downarrow&&\\ 0&\rightarrow&0&\rightarrow&C&\rightarrow&C’&&\end{array}$$
and
$$\begin{array}{ccccccccc}&&B/{\rm im}(\Delta)&\rightarrow&B’&\rightarrow&0&\rightarrow&0 \\ &&\downarrow&&\downarrow&&\downarrow&&\\ 0&\rightarrow&C&\rightarrow&C’&\rightarrow&0&&\end{array},$$ where the arrows are maps derived from the original problem with $g^{-1}:C\rightarrow C’.$
