Let $R$ be a ring, and consider an extension $\mathscr E : 0 \to N \to E \to M \to 0$ of $R$-modules.
If $h : N \to N'$ is a homomorphism, we can form the pushout
$h_*(\mathscr E ) : 0 \to N' \to E' \to M \to 0 $.
On the other hand, $h$ induces a homomorphism $h_* : Ext^1_R(M, N) \to Ext^1_R(M, N')$.
Under this homomorphism, the class $[\mathscr E ] \in Ext^1_R(M, N)$ is mapped to the class $[h_*(\mathscr E)] \in Ext^1_R(M, N')$.
In the above situation, if $ h : N \to N'$ is an isomorphism, show that $E \cong E'$ as R-modules.
Any idea how to solve this?
I think here I should use the fact that $Ext^1_R(M, N) \cong YExt^1_R(M, N)$, where $YExt^1_R(M, N)$ is what in my course was defined as the set of isomorphism clases in $YExt(M, N)$ and $YExt(M, N)$ is the category whose objects are extensions of the form $0 \to N \to E \to M \to 0$ and the morphisms are commutative diagrams
$0 \to N \to E \to M \to 0$
$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\downarrow{id}\text{ }$ $\text{ }\downarrow{f}\text{ }$ $\text{ }\downarrow{id}\text{ }$
$0 \to N \to E' \to M \to 0$
By definition, the pushout is related to the original extension by the diagram:
$0 \to N \xrightarrow{a} E \xrightarrow{b} M \to 0$
$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\downarrow{h}\text{ }$ $\text{ }\downarrow{\theta}\text{ }$ $\text{ }\downarrow{id}\text{ }$
$0 \to N' \xrightarrow{\alpha} E' \xrightarrow{\beta} M \to 0$
with $E'=h_*E=N'\oplus E/\{(h(n),-a(n))|n \in \Bbb N\}$
$ \alpha(n')=(n',0)$
$ \beta[n',x]=b(x)$
h is an isomorphism by assumption
I want to prove that $\theta$ is an isomorphism of modules But I have no idea what to do with this
Hint from the comments: Check that the diagram commutes using the universal property of the pushout, and then apply the Five Lemma.