Let's say we have the following diagram
$$\require{AMScd}\begin{CD}
0 @>>> A @>>> B @>>> C @>>> 0\\
{} @V{\alpha}VV @V{\beta}VV @V{\gamma}VV {} \\
0 @>>> A' @>>> B' @>>> C' @>>> 0
\end{CD}$$
where the top and bottom rows are short exact, so if $f: A \rightarrow B$ and $f': A' \rightarrow B'$ are injectives and $g:B \rightarrow C$, $g': B' \rightarrow C'$ surjectives, then $\text{Im}(f)= \text{ker}(g)$ and $\text{Im}(f')= \text{ker}(g')$. The short five lemma says that if the diagram is commutative and $\alpha$ and $\gamma$ are modules isomorphisms, then $\beta$ is an isomorphism.
Is it true that if $A \simeq A'$ and $C \simeq C'$ then there's an isomorphism $\beta :B \rightarrow B'$ such that the diagram is commutative? The hint is: let $A = A' = \mathbb{Z}_2$, $C = C' = \mathbb{Z}_2 \oplus \mathbb{Z}_2$ and $B = B' = \mathbb{Z}_4 \oplus \mathbb{Z}_2$ with $\alpha = \gamma = id$ and
$$g(a \text{ mod }4, b\text{ mod }2) = (a\text{ mod }2,b\text{ mod }2) \\
g'(a\text{ mod }4, b\text{ mod }2) = (b\text{ mod }2,a\text{ mod }2).$$
Find two injective morphisms $f,f': \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \oplus \mathbb{Z}_2$ such that the top and bottom row are exact but there's no isomorphism $\beta: \mathbb{Z}_4 \oplus \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \oplus \mathbb{Z}_2$ such that the diagram is commutative.
So for my proof, I started by finding
$$\text{ker}(g) = \text{ker}(g') = \left\{(0,0), (2,0) \right\},$$
here I mean $(0,0) = ([0],[0])$ the corresponding classes. So, the functions $f,f'$ that satisfies the conditions are $f=f'$ such that $f(0) = (0,0), f(1) = (2,0)$ (because $\text{Im}(f)=\text{ker}(g)$). Now, suppose that there is an isomorphism $\beta: \mathbb{Z}_4 \oplus \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \oplus \mathbb{Z}_2$ such that the diagram is commutative. Then $\beta \circ f = f' \circ \alpha$ i.e., $\beta \circ f = f$. So I should prove that $g=g' \circ \beta$ doesn't hold. Now my teacher said that since there're only $4$ possible $\beta$ isomorphisms, this can be made by hand (it is easy if $\beta = id$ because $g \neq g'$). My question is, is there a way to complete the proof without checking the condition for all possible isomorphisms $\beta$? Thanks
Best Answer
Two much work.
Simply observe that there are pairs of short exact sequences such as $$ 0\to \Bbb Z_2\to\Bbb Z_4\to \Bbb Z_2\to 0$$ and $$ 0\to \Bbb Z_2\to\Bbb Z_2\oplus \Bbb Z_2\to \Bbb Z_2\to 0$$ where the middle terms are not isomorphic.