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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?

CAA
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    Do you mean $X'\cong X$? If so, there's a nice result called "the five lemma" that tells you that $Y'\cong Y$ as well (assuming (1) and (2)). – Arthur Jul 10 '15 at 20:55
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    The Five Lemma requires that there exists a a diagram morphism between (1) and (2). If you don't assume that, then (1) and (2) (or (1) and (3)) need not be equivalent. – Batominovski Jul 10 '15 at 20:59
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    You need to clarify what the arrows are in these sequences. (PS: ".. is correct" is not a very precise statement: what you mean here is something like "if diagram (1) and diagram (2) are exact sequences then so is diagram (3)" and that requires you to say more about the arrows, e.g, to postulate the commutative properties of the 5 lemma.) – Rob Arthan Jul 10 '15 at 21:00
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    For example, there are two possible ways to define the short exact sequence $0\to\mathbb{Z}\to\mathbb{Z}\oplus \left(\mathbb{Z}/2\mathbb{Z}\right)^\omega \to (\mathbb{Z}/2\mathbb{Z})^\omega\to 0$ such that the resulting diagrams are not equivalent. One is the usual splitting sequence. – Batominovski Jul 10 '15 at 21:01
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    The other is given by $\mathbb{Z}\to\mathbb{Z}\oplus\left(\mathbb{Z}/2\mathbb{Z}\right)^\omega$ sending $x\in\mathbb{Z}$ to $\left(2x,\bar{0},\bar{0},\ldots\right)$, and $\mathbb{Z}\oplus\left(\mathbb{Z}/2\mathbb{Z}\right)^\omega\to\left(\mathbb{Z}/2\mathbb{Z}\right)^\omega$ sending $\left(x,\bar{y}_1,\bar{y}_2,\ldots\right)$ to $\left(\bar{x},\bar{y}_1,\bar{y}_2,\ldots\right)$. – Batominovski Jul 10 '15 at 21:04
  • @Batominovski Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Aug 22 '15 at 19:22

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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.

Batominovski
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