I am learning about $Ext^1(A,C)$ and how it forms a group under '$+$', the Baer sum and I am clearly missing the point somewhere. So, let us suppose for simplicity that $Ext^1(A,C)\cong\mathbb{Z}/3$. So we have three distinct extensions (up to congruence):
$I$: $0\to C\to A\oplus C\to A\to 0$;
$E$: $0\to C\stackrel{i}{\to}B\stackrel{p}{\to}A\to 0$;
$E^\prime$: $0\to C\stackrel{j}{\to}B^\prime\stackrel{q}{\to}A\to 0$.
Here we have $E+I\equiv E$ and $E^\prime+I\equiv E^\prime$ where '$\equiv$' represents congruence. Now, apparently we have the inverse extension'$-E$', which is the extension $0\to C\stackrel{i}{\to}B\stackrel{-p}{\to}A\to 0$ and $E+(-E)\equiv I$. So surely $-E\equiv E^\prime$ which implies $B\cong B^\prime$ which implies $E\equiv E^\prime$ which implies $Ext^1(A,C)\cong\mathbb{Z}/2$. Clearly I am making a mistake somewhere but I am not sure where! Any help would be much appreciated. Thanks.
The extension is characterized by the triple $(i,B,p)$ (not by the isomorphism class of $B$ only) which is different of $(i,B,-p)$.
The equivalence of extension is defined by this commutative diagram:https://en.wikipedia.org/wiki/Ext_functor#Equivalence_of_extensions not only by the isomorphism between $B$ and $B'$.