On induced maps in group homology and Künneth formula

Question

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)

Answer 1

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, Homology Homotopy Appl. 8, No. 2, 91-103 (2006). ZBL1107.55003.