If you search the site for this proof, you will find duplicates, however they are hard to understand. In other words they brush by the most critical points of the proof as if they were not worth expanding upon.
Similarly does Lang's Algebra do this on page 123 in the proof of Prop 2.1.
I will quote the book to see if others can figure out what I am not getting.
A sequence
$X' \xrightarrow{\lambda} X \to X'' \to 0$
is exact if and only if the sequence
$\text{Hom}(X', Y) \leftarrow \text{Hom}(X, Y) \leftarrow \text{Hom}(X'', Y) \leftarrow 0$
is exact for all $A$-modules $Y$.
I understand how to prove injectivity on the right, so I'm stuck on exactness at $\text{Hom}(X, Y)$ given exactness at $X$ of the first sequence.
I am having trouble showing the specific direction $\ker \lambda^* \subset \operatorname{im} \gamma^*$ where for example $\lambda^* = \text{Hom}(\lambda, Y)$ and $X \xrightarrow{\gamma} X''$ in the first sequence is the label given to that previously unlabled arrow that follows $\lambda$.
On that particular part, the book says:
Let $g \in \ker \gamma^*$ i.e. $g \circ \gamma = 0$. In other words:
Then $g$ vanishes on the image of $\lambda$. Hence we can factor $g$ through the factor module:
$$
\require{AMScd}
\begin{CD}
X @>>> X/\operatorname{im} \lambda \\
@VgVV @VVV \\
Y @= Y
\end{CD}
$$
Hence we can factor $g$ through $X''$, thereby showing that the kernel of
$\text{Hom}(X', Y) \leftarrow \text{Hom}(X, Y)$
is contained in the image of …
Firstly, what isomorphism theorem is being used (the second time isomorphism is mentioned above), and secondly how does factoring through $X''$ come about?
Best Answer
Let $$X' \xrightarrow{\lambda} X \xrightarrow{\gamma} X'' \to 0$$ be an exact sequence of $A$-modules.
Let $Y$ be an $A$-module. You say you are having trouble showing that if $$\gamma^* : \text{Hom}(X'', Y) \to \text{Hom}(X, Y)$$ $$\lambda^* : \text{Hom}(X, Y) \to \text{Hom}(X', Y)$$ are the induced maps, then $\ker (\lambda^*) \subset \text{Im}( \gamma^*)$. So in the great mathematical tradition I will show that and leave everything else as an exercise.
Suppose that $g \in \ker \lambda^*$ then $g \circ \lambda = 0$ hence $g$ vanishes on $ \text{Im}(\lambda) = \ker(\gamma)$. In particular we may factor the map $g:X \to Y$ through $$X \xrightarrow{\gamma} X'' \xrightarrow{f} Y$$ where we implicitly use $X/\text{Im}(\gamma) \cong X''$ and the second map we denote by $f$. In particular $\gamma^*(f) = f \circ \gamma = g$. Thus $g \in \text{Im}(\gamma^*)$.
NB: In the course of asking this I think that you mixed up some of your $\lambda^*$ and $\gamma^*$.