I have some problem calculating the value of some specific (but quite common) induced maps I stumbled on while reading some papers on group (co)homology and I would like to know if there are general tricks or routines to avoid calculations over the complexes or, if not, which calculations are customary. But let's first fix the notation.



> 
> Let $H$ and $H'$ be groups, let $M$ be an $H$-module and $M'$ be an $H'$-module. If $\phi:H\rightarrow H'$ is a homomorphism and $\psi:M\rightarrow M'$ is an $H$-module homomorphism with $M'$ regarded as an $H$-module via $\phi$, then it induces a map $(\phi,\psi)\_\*:H\_i(H,M)\rightarrow H\_i(H',M')$ and this is done taking two projective resolutions $\textbf{P}$ and $\textbf{P'}$ of $\mathbb{Z}$ over $\mathbb{Z}H$ and $\mathbb{Z}H'$, respectively, making the latter a resolution over $\mathbb{Z}H'$, noticing that it is still acyclic and using this and the projectivity of $\textbf{P}$ to build the map we are looking for. If $M=M'$ and $\psi=Id\_{M'}$, it is customary to write $(\phi,\psi)\_\*$ as $\phi\_\*$.
> 
> 
> 


Now for the specific cases.


1. Let $G=H$, $G'=H\times H$, $M=K=M'$ for a field $K$ on which $G$ acts trivially. Calculate $\phi\_\*$ where $\phi$ is either $\alpha:g\rightarrow (g,1)$ or $\beta:g\rightarrow (1,g)$
or $\gamma:g\rightarrow (g,g)$. In this case a Künneth formula states that
$$H\_n(G',M)=\bigoplus\_{p=0}^n H\_p(G,M)\otimes\_M H\_{n-p}(G,M).$$ How can one get that the $(H\_0(G,M)\otimes\_M H\_n(G,M)\oplus H\_n(G,M)\otimes\_M H\_0(G,M))$-components of $\alpha\_\*(h)$, $\beta\_\*(h)$ and $\gamma\_\*(h)$ for an $h$ in $H\_n(G,M)$ are, resp., $h\otimes 1$, $1\otimes h$ and $h\otimes 1+1\otimes h$?
2. More generally, let $G$, $G'$, $\alpha$, $\beta$ and $\gamma$ be the same as in point 1., $M$ and $N$ be $G$-modules, $M'=M\otimes\_\mathbb{Z} N$ be a $G'$-module by the action $(g,h)\cdot(m\otimes\_\mathbb{Z} n)=gm\otimes\_\mathbb{Z} hn$ and let $\psi:M\rightarrow M'$ be a suitable $G$-module homomorphism (Doubt: For $\alpha$, $\psi$ could just send $m\in M$ to $m\otimes\_\mathbb{Z}\overline{n}$ for a fixed $\overline{n}\in N$ and it seems the same holds for $\gamma$ if $G$ acts trivially on $N$; but for $\beta$?). What is the best way to find the $(H\_0(G,M)\otimes\_M H\_n(G,M)\oplus H\_n(G,M)\otimes\_M H\_0(G,M))$-components of $(\alpha,\psi)\_\*(h)$, $(\beta,\psi)\_\*(h)$ and $(\gamma,\psi)\_\*(h)$ for an $h$ in $H\_n(G,M)$, provided that a formula like the following $$H\_n(G',M\otimes\_\mathbb{Z} N)=\bigoplus\_{p=0}^n H\_p(G,M)\otimes\_\mathbb{Z} H\_{n-p}(G,M')$$ still holds? (This should be possible by choosing $M$ and $H\_i(G,M')$ to be $\mathbb{Z}$-free)



On induced maps in group homology and Künneth formula
1. The Kunneth formula is natural for products of homomorphisms. Let $\iota:\{1\}\to H$ be the inclusion of the identity element, and consider $\mathrm{Id}\times \iota:H\times\{1\}\to H\times H$ which can be identified with your map $\alpha$. Applying the Kunneth formula, we see that an element $h\in H\_n(H\times\{1\};M)\cong H\_n(H;M)\otimes H\_0(\{1\};M)$ maps to $h\times 1$ under $\alpha\_\*$. Similarly for $\beta\_\*$.


For the diagonal homomorphism $\gamma:H\to H\times H$, we can use functoriality of Kunneth for the projections $p\_1:H\times H\to H\times\{1\}$ and $p\_2:H\times H\to \{1\}\times H$, and observe that $(p\_1)\_\*$ maps an element in $H\_n(H\times H;M)$ to its $H\_n(H;M)\otimes H\_0(H;M)$ component, and $(p\_2)\_\*$ maps an element in $H\_n(H\times H;M)$ to its $H\_0(H;M)\otimes H\_n(H;M)$ component. Then $p\_1\circ \gamma$ and $p\_2\circ \gamma$ are both idetified with $\mathrm{Id}\_H$.


2. You might be able to apply a similar reasoning with twisted coefficients. A reference for the Kunneth formula in this case is


*Greenblatt, Robert*, [**Homology with local coefficients and characteristic classes**](http://dx.doi.org/10.4310/HHA.2006.v8.n2.a5), Homology Homotopy Appl. 8, No. 2, 91-103 (2006). [ZBL1107.55003](https://zbmath.org/?q=an:1107.55003).


