On the wikipedia page for the Ext functor, they say that one can equip the graded abelian group $\operatorname{Ext}^*:=\bigoplus_{i=0}^{\infty}\operatorname{Ext}^i(A,A)$ with the structure of a ring (for $A$ an $R$-module for some ring $R$) by identifying $\operatorname{Ext}^i(A,B)$ with the group of chain homotopy classes of chain maps $P\rightarrow Q[i]$ (where $P,Q$ are projective resolutions of $A,B$) where the Yoneda product $\operatorname{Ext}^i(A,B)\otimes \operatorname{Ext}^j(B,C)\rightarrow \operatorname{Ext}^{i+j}(A,C)$ is given by composing chain maps.
Apparently this identification (of $\operatorname{Ext}^i(A,B)$ with these chain maps) can be made by using the definition of Ext given by $\operatorname{Ext}^i(A,B)=H^i (\operatorname{Tot}^{\prod}(\operatorname{Hom}(P_{\bullet},Q_{\bullet}))$ (which is the one I was taught) and using that composition of homomorphisms $\circ :\operatorname{Hom}(A,B)\otimes \operatorname{Hom}(B,C)\rightarrow \operatorname{Hom}(A,C)$ induces a well defined map $\operatorname{Ext}^i(A,B)\otimes \operatorname{Ext}^j(B,C)\rightarrow \operatorname{Ext}^{i+j}(A,C)$.
Could someone please explain to me how composition of morphisms induces this map and also how the identification of $\operatorname{Ext}^i$ with the chain maps between resolutions is made? I've read the wikipedia page for the Yoneda product and looked on the internet and in Weibel but they use a different identification in that case which isn't useful to me. Thanks!
Best Answer
$\newcommand{\Ext}{\operatorname{Ext}}$It is a bit easier to work with $\Ext^*(A,B)$ as computed by a projective resolution of $A$, say $P_*:\cdots\to P_n \to\cdots\to P_0$, so that an element of $\Ext^n$ is a linear map $f: P_n\to B$ such that $fd=0$. Now suppose that you want to define the composition of this with a map in $\Ext^*(B,C)$. Pick a projective resolution $Q_*$ of $B$, and a cocycle $g: Q_m\to C$ in the second $\Ext$-group.
You can consider the map $f:P_n\to B$ and lift it through $Q_0\to B$ since this last map is onto. Now the usual yoga of homological algebra allows you to lift this all the way until you get a map $f_m: P_{m+n}\to Q_m$. Now you can take the composite $gf_m : P_{m+n}\to C$ and obtain an element in $\Ext^*(A,C)$, and this is the Yoneda product $f\smile g$, probably up to a sign.