Intuition behind $Ext^1(A,\,C)$ So I recently asked a question concering $Ext^1(A,\,C)$ regarding the connection between isomorphism and the congruence '$\equiv$' (Where am I making a mistake with $Ext^1(A,C)$?). Suppose, for simplicity that $Ext^1(A,\,C)\cong\mathbb{Z}/3$ and the non trivial extensions are given by:
$E$: $0\to C\to B\to A\to 0$;
$E^\prime$: $0\to C\to B^\prime\to 0$.
It was pointed out to me that $B\cong B^\prime$ without $E\equiv E^\prime$ and this seems quite odd to me, intuitively. I was wondering if anyone could provide an intuition about what is going on, and want each class of extensions is actually telling us. Thanks!
 A: When computing $Ext^1(A,C)$, you are not only interested in how many $B$ can be fit into an exact sequence
$$0\to C\to B\to A\to 0$$
but also, for a fixed $B$, how many distinct ways can you construct such a sequence. Of course, distinct means "distinct up to equivalence" as linked to in your previous question. That is, there is an isomorphism $\phi$ such that the following diagram commutes:
$$
\begin{array}{ccccccccc}
0&\longrightarrow&C&\longrightarrow&B&\longrightarrow&A&\longrightarrow&0\\
&&\downarrow^{id}&&\downarrow^{\phi}&&\downarrow^{id}&&\\
0&\longrightarrow&C&\longrightarrow&B&\longrightarrow&A&\longrightarrow&0
\end{array}
$$
Morally, you should think of two sequences as being equivalent if you can get from one to the other via an automorphism of $B$.
It is instructive to work out the case $A=C=\mathbb{Z}/3$. In this case, there are only two extensions of $\mathbb{Z}/3$ by $\mathbb{Z}/3$, namely $\mathbb{Z}/3\times\mathbb{Z}/3$ and $\mathbb{Z}/9$. However, $Ext^1(\mathbb{Z}/3,\mathbb{Z}/3)=\mathbb{Z}/3$. To see this, note that there are precisely two embeddings $$\iota_1:\mathbb{Z}/3\to\mathbb{Z}/9\;\;\;\mbox{and}\;\;\; \iota_2:\mathbb{Z}/3\to\mathbb{Z}/9$$ given by $\iota_1(1)=3$ and $\iota_2(1)=6$, respectively. Similarly, there are two projections 
$$\pi_1:\mathbb{Z}/9\to\mathbb{Z}/3\;\;\;\mbox{and}\;\;\; \iota_2:\mathbb{Z}/9\to\mathbb{Z}/3$$
given by $\pi_1(1)=1$ and $\pi_2(1)=2$. 
The two nontrivial elements of $Ext^1(\mathbb{Z}/3,\mathbb{Z}/3)$ are
$$
0\to\mathbb{Z}/3\;\longrightarrow^{\iota_1}\;\mathbb{Z}/9\;\longrightarrow^{\pi_1}\;\mathbb{Z}/3\to 0
$$
and
$$
0\to\mathbb{Z}/3\;\longrightarrow^{\iota_1}\;\mathbb{Z}/9\;\longrightarrow^{\pi_2}\;\mathbb{Z}/3\to 0
$$
and all remaining diagrams are equivalent to one of these via the automorphism $\phi:\mathbb{Z}/9\to\mathbb{Z}/9$ given by $\phi(1)=-1$. Indeed, it is enough to check the commutativity of the following diagrams:
$$
\begin{array}{ccccccccc}
0&\longrightarrow&\mathbb{Z}/3&\longrightarrow^{\iota_1}&\mathbb{Z}/9&\longrightarrow^{\pi_1}&\mathbb{Z}/3&\longrightarrow&0\\
&&\downarrow^{id}&&\downarrow^{\phi}&&\downarrow^{id}&&\\
0&\longrightarrow&\mathbb{Z}/3&\longrightarrow^{\iota_2}&\mathbb{Z}/9&\longrightarrow^{\pi_2}&\mathbb{Z}/3&\longrightarrow&0
\end{array}
$$
and
$$
\begin{array}{ccccccccc}
0&\longrightarrow&\mathbb{Z}/3&\longrightarrow^{\iota_1}&\mathbb{Z}/9&\longrightarrow^{\pi_2}&\mathbb{Z}/3&\longrightarrow&0\\
&&\downarrow^{id}&&\downarrow^{\phi}&&\downarrow^{id}&&\\
0&\longrightarrow&\mathbb{Z}/3&\longrightarrow^{\iota_2}&\mathbb{Z}/9&\longrightarrow^{\pi_1}&\mathbb{Z}/3&\longrightarrow&0
\end{array}
$$
which is a straightforward exercise.
