How do you explain why the arguments of $\operatorname{Ext}^1(A,B)$ aren't "backwards"?

Question

For objects $A$ and $B$ In an abelian category, $\operatorname{Hom}(A,B)$ is the group of morphisms $$A \longrightarrow B\,.$$ Now $\operatorname{Ext}^1(A,B)$, the derived functor of $\operatorname{Hom},$ can be thought of as the group of extensions of $A$ by $B$, so short exact sequences (up to equivalence) of the form $$B \hookrightarrow \_\_ \twoheadrightarrow A\,.$$ It looks like $A$ and $B$ have switched places. If we insist that arrows go from left to right, then naïvely thinking about it $\operatorname{Ext}^1(A,B)$ should be sequences that start with $A$ on the left and end with $B$ on the right, like with $\operatorname{Hom}$. Now $A$ and $B$ aren't really backwards, and this makes sense somehow, but I've never thought much of it until seeing fellow students make the naïve mistake I mentioned. How can you briefly explain to someone that this isn't a notational quirk, but, relating $\operatorname{Ext}$ to $\operatorname{Hom}$, this actually makes sense?

Answer 1

The machinery of derived categories makes this much more intuitive. In particular, an element of $\operatorname{Ext}^1(A,B)$ can be identified with a morphism $A\to B[1]$ in the derived category (where $A$ and $B$ are considered as objects in the derived category by treating them as chain complexes concentrated in degree $0$). Taking the fiber of this morphism, we get an object $C$ in the derived category and an exact triangle $$B\to C\to A\to B[1].$$ Looking at the associated long exact sequence of homology objects, since $B$ and $A$ have homology concentrated in degree $0$, so does $C$, and the long exact sequence on homology just turns into a short exact sequence $$0\to B\to C\to A\to 0$$ where now $C$ represents the $0$th homology of the derived object $C$ we had before. That's an extension of $A$ by $B$! Conversely, given such an extension, we can consider $A,B,$ and $C$ as objects in the derived category and get an exact triangle $B\to C\to A\to B[1]$ as above and in particular we get a derived morphism $A\to B[1]$, and these two constructions can be shown to be inverse.

Or, without the language of derived categories, we should think of an extension $$0\to B\to C\to A\to 0$$ as "going from $A$ to $B$" since when you look at long exact sequences associated to this short exact sequence, the connecting homomorphisms of the long exact sequence go from $A$ to $B$ (with a degree shift). Those connecting homomorphisms correspond exactly to the morphism $A\to B[1]$ in the derived category.

Answer 2

Just use the long exact sequence. $Ext(A,B)$ extends the sequence $0\to Hom(A,B)\to Hom(A,B')\to Hom(A,B'')$ when $B\to B'\to B''$ is short exact. More broadly it is the derived functor of $Hom(A,-)$ or of $Hom(-,B)$, and more abstractly it's the homs from $A$ to the suspension of $B$ in the derived category. All in all, unless the only way you've seen $Ext$ is as a set of extensions, which seems pedagogically rare, then there's plenty of reason to write the arguments in the correct order.