# Understanding the Yoneda product defined in terms of morphisms of projective resolutions.

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 $$\operatorname{Ext}^i(A,B)\otimes \operatorname{Ext}^j(B,C)\rightarrow \operatorname{Ext}^i(A,C)$$ 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!

$$\newcommand{\Ext}{\operatorname{Ext}}$$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.