Let $\mathcal A$ be an Abelian category. We can assume it has enough injectives. Let $\operatorname{SES}\_{\mathcal A}$ denote the category, of which objects are short exact sequences in $\mathcal A$, and morphisms are triples $(f,f',f'')$ fitting in the diagram
$$\require{AMScd}\begin{CD}0@>>>E'@>>>E@>>>E''@>>>0\\{}@Vf'VV@VfVV@Vf''VV\\0@>>>F'@>>>F@>>>F''@>>>0.\end{CD}$$


We call an object $[I]\in\operatorname{SES}\_{\mathcal A}$ *termwise injective* if it is a short exact sequence
$$\require{AMScd}\begin{CD}0@>>>I'@>>>I@>>>I''@>>>0\end{CD}$$
and $I,I',I''$ are injective objects in $\mathcal A$.


Since $\mathcal A$ has enough injectives, for any object $[E]\in\operatorname{SES}\_{\mathcal A}$, it has a termwise injective resolution
$$\require{AMScd}\begin{CD}0@>>>[E]@>>>[I^0]@>>>[I^1]@>>>\cdots\end{CD}$$


Now I want to define $\operatorname{Ext}^2([E],[E])$ as the homology group at middle of the following sequence
$$\require{AMScd}\begin{CD}\prod\_k\operatorname{Hom}([I^k],[I^{k+1}])@>>>\prod\_k\operatorname{Hom}([I^k],[I^{k+2}])@>>> \prod\_k\operatorname{Hom}([I^k],[I^{k+3}])\end{CD}. $$


Here I want to ask whether the construction above is independent on the resolution.



Can we define $\operatorname{Ext}$ groups in the category of short exact sequences?
Unfortunately, short exact sequences don't form an abelian category; the snake lemma shows you where the problem is. However, they do form an exact category, in which the exact sequences are the diagrams with nine terms and exact rows and columns. Ext can be defined in any exact category, via the Yoneda definition. The split short exact sequences of injectives are injective objects in this exact category (this needs checking by a diagram chase which I haven't written out, so buyer beware). So such resolutions should compute Ext in the normal way. In any case, this does isolate what needs to be checked.


