Suppose I have a short exact sequence in an abelian category (say abelian groups for simplicity)
$$0 \to B \to X \to A \to 0.$$
If I apply $\operatorname{Ext}^*(C, \bullet)$ to this sequence, I get a long exact sequence. Let's take $C = \mathbb{Z}/2\mathbb{Z}$ for simplicity:
$$0 \to \operatorname{Hom}(\mathbb{Z}/2\mathbb{Z}, B) \to \operatorname{Hom}(\mathbb{Z}/2\mathbb{Z}, X) \to \operatorname{Hom}(\mathbb{Z}/2\mathbb{Z}, A) \to \operatorname{Ext}^1(\mathbb{Z}/2\mathbb{Z}, B)$$
$$ \to \operatorname{Ext}^1(\mathbb{Z}/2\mathbb{Z}, X) \to \operatorname{Ext}^1(\mathbb{Z}/2\mathbb{Z}, A) \to 0.$$
I can figure out that $\operatorname{Ext}^1(\mathbb{Z}/2\mathbb{Z}, A) \cong A/2A$ by looking at a projective resolution of $\mathbb{Z}/2\mathbb{Z}$ and applying $\operatorname{Ext}^*(\bullet, A)$. So my exact sequence looks like
$$0 \to \operatorname{Hom}(\mathbb{Z}/2\mathbb{Z}, B) \to \operatorname{Hom}(\mathbb{Z}/2\mathbb{Z}, X) \to \operatorname{Hom}(\mathbb{Z}/2\mathbb{Z}, A) \overset{f}{\to} B/2B$$
$$ \overset{g}{\to} X/2X \overset{h}{\to} A/2A \to 0.$$
My Question: How can I understand the maps $f$, $g$, and $h$? I can understand the analogous maps when I apply the contravariant functor $\operatorname{Ext}^*(\bullet, C)$, because I can find projective resolutions for $B$ and $A$, apply the Horseshoe Lemma and get a short exact sequence of projective resolutions, and just look at where everything goes. But in order to do this for the covariant $\operatorname{Ext}$ functor, I'd have to construct injective resolutions, which I don't know how to do concretely. So how can I figure out what the maps $f$, $g$, and $h$ should be?
Abstract Algebra – Understanding Maps in Long Exact Sequence of Ext
abstract-algebrahomological-algebra
Best Answer
As you said, you can compute $\operatorname{Ext}(\mathbb{Z}/2\mathbb{Z},\cdot)$ using a projective resolution of $\mathbb{Z}/2\mathbb{Z}$. But you can also use this projective resolution to understand what the maps $f,g,h$ are ! You will find out that the maps $g$ and $h$ are the obvious one, namely a map $X\rightarrow Y$ induces a map $X/2X\rightarrow Y/2Y$, and those are the ones that arises in the long sequence of $\operatorname{Ext}$.
The map $f$ is a connecting homomorphism and as such it is not always very obvious. As often it comes from the snake lemma, but here you can draw the entire diagram to get everything :
$$\require{AMScd} \begin{CD} 0@>>>\operatorname{Hom}(\mathbb{Z}/2\mathbb{Z},B)@>>>\operatorname{Hom}(\mathbb{Z}/2\mathbb{Z},X)@>>>\operatorname{Hom}(\mathbb{Z}/2\mathbb{Z},A)\\ @.@VVV@VVV@VVV\\ 0@>>>\operatorname{Hom}(\mathbb{Z},B)@>>>\operatorname{Hom}(\mathbb{Z},X)@>>>\operatorname{Hom}(\mathbb{Z},A)@>>>0\\ @.@V{\times 2}VV@V{\times 2}VV@V{\times 2}VV\\ 0@>>>\operatorname{Hom}(\mathbb{Z},B)@>>>\operatorname{Hom}(\mathbb{Z},X)@>>>\operatorname{Hom}(\mathbb{Z},A)@>>>0\\ @.@VVV@VVV@VVV\\ @.\operatorname{Ext}(\mathbb{Z}/2\mathbb{Z},B)@>>>\operatorname{Ext}(\mathbb{Z}/2\mathbb{Z},X)@>>>\operatorname{Ext}(\mathbb{Z}/2\mathbb{Z},A)@>>>0 \end{CD} $$ or using identifications $\operatorname{Hom}(\mathbb{Z},X)=X$ and $\operatorname{Hom}(\mathbb{Z}/2\mathbb{Z},X)=X[2]$, this diagram can be written : $$\require{AMScd} \begin{CD} 0@>>>B[2]@>>>X[2]@>>>A[2]\\ @.@VVV@VVV@VVV\\ 0@>>>B@>>>X@>>>A@>>>0\\ @.@V{\times 2}VV@V{\times 2}VV@V{\times 2}VV\\ 0@>>>B@>>>X@>>>A@>>>0\\ @.@VVV@VVV@VVV\\ @.B/2B@>>>X/2X@>>>A/2A@>>>0 \end{CD} $$ with the natural maps. The connecting homomorphism from the snake lemma gives a long exact sequence $$0\rightarrow B[2]\rightarrow X[2]\rightarrow A[2]\overset{f}\rightarrow B/2B\rightarrow X/2X\rightarrow A/2A\rightarrow 0$$ Finally the map $f$ is the following : take an element $a$ of order 2 in $A$, lift it in $X$ to get an element $x$. Multiply $x$ by 2, it will land in $B$ and take its class modulo $2B$.