Show that the maps \(G \xrightarrow{n} G\) and \(H \xrightarrow{n} H\) multiplying each element by the integer \(n\) induce multiplication by \(n\) in \(\operatorname{Ext}(H, G)\).
Solution
Choose a free resolution \[ \cdots\longrightarrow F_1\xrightarrow{d_1}F_0\longrightarrow H\longrightarrow0. \] Then \[ \operatorname{Ext}(H,G)=H^1(\operatorname{Hom}(F_\bullet,G)). \]
1. Multiplication by \(n\) on \(G\) induces multiplication by \(n\) on \(\operatorname{Ext}(H,G)\).
The map \[ [n]_G:G\to G,\qquad g\mapsto ng \] induces on every cochain group the map \[ \operatorname{Hom}(F_i,G)\to\operatorname{Hom}(F_i,G), \qquad \varphi\mapsto [n]_G\circ\varphi=n\varphi. \] Thus the induced cochain map is literally multiplication by \(n\) in every degree. Its map on cohomology, in particular on \(H^1=\operatorname{Ext}(H,G)\), is therefore multiplication by \(n\).
2. Multiplication by \(n\) on \(H\) also induces multiplication by \(n\) on \(\operatorname{Ext}(H,G)\).
The endomorphism \([n]_H:H\to H\) lifts to the chain endomorphism \([n]_{F_i}:F_i\to F_i\) given by multiplication by \(n\) in each degree, since all differentials are homomorphisms and hence commute with multiplication by \(n\). Because \(\operatorname{Ext}\) is contravariant in its first variable, the resulting cochain map is precomposition: \[ \varphi\longmapsto \varphi\circ[n]_{F_i}=n\varphi. \] Again this is multiplication by \(n\) on every cochain group, so it induces multiplication by \(n\) on cohomology.
Hence both endomorphisms specified in the problem induce \[ \boxed{n\cdot\operatorname{id}_{\operatorname{Ext}(H,G)}}. \]