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Suppose we have an exact sequence of $R$-modules

\begin{array}{ccccccccc} 0 & \longrightarrow & L & \overset{f}{\longrightarrow} & M & \overset{g}{\longrightarrow} & E & \longrightarrow & 0\\ \end{array} Let $\phi : L \to L'$ be a homomorphism of $R$-modules.

We know that there is a bijection between the ( classes of isomorphic ) extensions $e(E,L)$ and $\text{Ext}_1(E,L)$.

Moreover Ext$_1(-,-)$ is a functor contravariant in the first component and covariant in the second. My question is: what is the concrete morphism of extensions induced by $$\text{Ext}_1(E,\phi) : \text{Ext}_1(E,L) \to \text{Ext}_1(E,L')$$ ? i.e. what are the maps which define a commutative diagram

$$\begin{array} 00 & {\longrightarrow} & L & \stackrel{f}{\longrightarrow} & M \stackrel{g}{\longrightarrow} & E & {\longrightarrow} & 0\\ & & \downarrow{\alpha} & & \downarrow{\beta} & \downarrow{id} \\ 0 & {\longrightarrow} & L' & \stackrel{f'}{\longrightarrow} & M' \stackrel{g'}{\longrightarrow} & E & {\longrightarrow} & 0\\ \end{array} $$

?

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  • $\begingroup$ Ext classifying extensions should be thought of as a theorem, not a definition (in my opinion). The definition is that Ext is "derived Hom," and in particular the morphism you're asking about is a composition of a derived homomorphism with an ordinary homomorphism. $\endgroup$
    – Qiaochu Yuan
    Jun 4, 2014 at 19:59
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    $\begingroup$ This is explained nicely in Mac Lane's book Homology. $\endgroup$
    – Martin Brandenburg
    Jun 5, 2014 at 10:26

1 Answer 1

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You have to form the pushout of $f$ and $\alpha$ to obtain $f'$ and $\beta$, then show that $E$ will be a cokernel of $f'$.

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