Understanding $\operatorname{Ext}_R^1(M,A)$ as obstructions to lifting homomorphisms

Question

Let $R$ be a ring. Let's work in the category of left $R$-modules. There is a way of looking at $\operatorname{Ext}_R^1(M,A)$ as classifying extensions of $M$ by $A$ up to equivalence.

But, if we look at the $\operatorname{Ext}_R^1(M, -)$ as the first derived functor of $\hom(M, -)$, for every short exact sequence of $R$-modules $$0 \to A \to B \to C \to 0 $$ we get a long exact sequence of form $$0 \to \hom(M,A) \to \hom(M,B) \to \hom(M,C) \to \operatorname{Ext}_R^1(M,A) \to \operatorname{Ext}_R^1(M,B) \to \operatorname{Ext}_R^1(M,C) \to \operatorname{Ext}_R^2(M,A) \to \ldots $$

In this formulation, we could say that $\operatorname{Ext}_R^1(M,A)$ in a way measures failure to lift homomorphisms $M \to C$ to homomorphisms $M \to B$. This gives a connection between extensions of $M$ by $A$ and lifting of homomorphisms $M \to C$.

For an example, if we have no non-trivial extension of $M$ by $A$, then $\operatorname{Ext}_R^1(M,A) = 0$ and we can always lift homomorphisms $M \to B$ to homomorphisms $M \to C$.

My question is: what would be a more concrete way of connecting these two notions? If I have a homomorphism $M \to C$ that I can't lift to homomorphism $M \to B$, do I automatically get an extension of $M$ by $A$? Is this described in some book/set of lecture notes?

Answer 1

You can construct the connecting morphism $\hom(M,C) \to \operatorname{Ext}^1(M,A)$ explicitly in terms of the short exact sequence, and this should give you an idea of what's going on. Indeed, given a map $u:M\to C$, "plug it in" the diagram of the sequence and form the pullback $E$ of $B\to C\leftarrow M$. The kernel of the pullback is naturally isomorphic to $A$, so you obtain an extension $\xi(u):A\to E\to M$, which lives in $\operatorname{Ext}^1(M,A)$. Its class gives a representative of $\partial(u)$.

Claim We can lift $u : M\to C$ to a map $h:M\to B$ if and only if the class of $\xi(u)$ is trivial.

Proof Suppose first the map can be lifted, so there is (calling the maps of your sec $f,g$ as usual) a map $h:M\to B$ such that $gh = u$. Recall that $E= \{ (b,m) : gb = um\}$. Since $gh=u$, this is just the set of $(b,m)$ such that $b-hm\in \ker g$. Now $\ker g = \operatorname{im} f\simeq A$ and we get a map $t: E\to A$ that sends $(fa,m)$ to $fa-hm$. This splits the inclusion $\operatorname{im} f\to E$, so by the splitting lemma the extension $\xi(u)$ is trivial.

Conversely, suppose that the extension is trivial, so there is a map $k: \operatorname{im}f\oplus M\to E$ such that $k(fa,0) = (fa,0)$ and $\pi k(fa,m) = m$. This means the map is of the form $k(fa,0) = (fa+hm,m)$ for some map $h$. But since this lands in $E$, we get that $g(fa+hm) =ghm = um$, so $h$ is a lift.$\hspace{8 cm}\blacktriangleleft$

It is important that you observe that not every class in $\operatorname{Ext}^1(M,A)$ arises as such obstruction. This happens, by the LES, if and only if the map $\operatorname{Ext}^1(M,f)$ is injective.