Short exact sequence and extension Let $$0\rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0 ~~~~~(1)$$ be a short exact sequence of abelian groups. Suppose $$0\rightarrow X^{'} \rightarrow Y^{'} \rightarrow Z^{'} \rightarrow   0   ~~~~~~~~~(2)$$ be another short exact sequence of abelain groups with $X^{'}\cong X$ and $Z^{'}\cong Z$ as abelian groups. From (1) and (2) we can get another short exact sequence by using isomorphism $X^{'}\cong$ and $Z^{'}\cong Z$ $$0\rightarrow X\rightarrow Y^{'} \rightarrow Z \rightarrow 0 ~~~~~~(3).$$ My question here the sequence $(3)$ is correct or not?. If $(1)$ and $(3)$ are equivalent is it enough to say $(1), (2)$ and $(3)$ are equivalent?
 A: As requested by Julian Kuelshammer, I shall put my comment here so that the question can be marked as answered.  If we have two short exact sequences of modules over a ring $R$, say $0\to X\to Y\to Z\to 0$ and $0\to X'\to Y'\to Z'\to 0$, then it does not hold in general that $X\cong X'$ and $Z\cong Z'$ imply that the two short exact sequences are equivalent, even if $Y\cong Y'$ is also true.  In a situation where there exists a diagram morphism, i.e., $R$-module homomorphisms $X\to X'$, $Y\to Y'$, and $Z\to Z'$ such that $X\to X'$ and $Y\to Y'$ are isomorphisms and that
$$\begin{array}
\mathrm{0} &\to & X & \to & Y & \to & Z  &\to &0 \\
  &    & \downarrow & &\downarrow & & \downarrow & &&\\
\mathrm{0} &\to & X' & \to & Y' & \to & Z'  &\to &0  
\end{array}
$$
is commutative, then it is guaranteed by the Five Lemma that $Y\cong Y'$ as well as that the two short exact sequences are equivalent in the category of short exact sequences of $R$-modules.
However, the information that $X\cong X'$ and $Z\cong Z'$ alone, even if $Y\cong Y'$ also holds, does not suffice.  An example is when $R=\mathbb{Z}$, $X=X'=\mathbb{Z}$, $Y=Y'=\mathbb{Z}\oplus (\mathbb{Z}/2\mathbb{Z})^\omega$, and $Z=Z'=(\mathbb{Z}/2\mathbb{Z})^\omega$.  For $0\to X\to Y\to Z\to 0$, let $X\to Y$ and $Y\to Z$ be the usual canonical injection and the usual canonical projection, respectively.  Clearly, this short exact sequence splits.  For the short exact sequence $0\to X'\to Y'\to Z'\to 0$, let $X'\to Y'$ be the map sending $x\in\mathbb{Z}$ to $(2x,\bar{0},\bar{0},\ldots)$, while $Y'\to Z'$ is the map sending $\left(x,\bar{y}_1,\bar{y}_2,\ldots\right)$ to $\left(\bar{x},\bar{y}_1,\bar{y}_2,\ldots\right)$ for all $x,y_1,y_2,\ldots\in\mathbb{Z}$.  It is an easy exercise to show that the latter short exact sequence does not split.  Hence, $0\to X\to Y\to Z\to 0$ and $0\to X'\to Y'\to Z'\to 0$ are not equivalent.
