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My question is about Shapiro's lemma. Consider the isomorphism $\phi: H^n(G, Hom_{ZH}(ZG, A))\cong H^n(H,A)$ of shapiro's lemma. I would like to describe this via cochains.

So the obvious map is $\phi(f+B^n(G,Hom_{ZH}(ZG, A) ))(h_1,\cdots, h_n)=f(h_1,\cdots, h_n)(1)+B^n(H, A)$.

QUESTION: What is the inverse of this map, i.e how to get a map from $H^n(H,A)\to H^n(G, Hom_{ZH}(ZG, A))$, which is the inverse of the above map?

Of course there is a standard proof of Shapiro's lemma, but using this approach it seems it is not possible to describe the inverse, of course we know that an inverse exists. So a further question would be what are some other isomorphisms where it is not possible to describe an inverse

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Take a standard homogeneous $H$-resolution of $\mathbb{Z}$:

$$\ldots \to \mathbb{Z}[H^3] \to \mathbb{Z}[H^2] \to \mathbb{Z}H \to \mathbb{Z} \to 0 \qquad (1)$$

where $\phi_n: \mathbb{Z}[H^{n+1}] \to \mathbb{Z}[H^n]$ for $n \geq 0$ takes $(h_0, \ldots, h_n)$ to $\sum_{i = 0}^n (-1)^i (h_0, \ldots, \hat{h_i}, \ldots, h_n)$. The cohomology of the induced complex $\hom_{\mathbb{Z}H}(\mathbb{Z}[H^n], A)$ gives the cochain description of $H^n(H, A)$.

The inverse map you want to describe at the cochain level is obtained by applying the cohomology functor to an appropriate map of cochain complexes

$$\hom_{\mathbb{Z}H}(\mathbb{Z}[H^n], A) \to \hom_{\mathbb{Z}G}(\mathbb{Z}[G^n], \hom_{\mathbb{Z}H}(\mathbb{Z}G, A)).$$

There is a canonical isomorphism $\hom_{\mathbb{Z}G}(\mathbb{Z}[G^n], \hom_{\mathbb{Z}H}(\mathbb{Z}G, A)) \cong \hom_{\mathbb{Z}H}(\mathbb{Z}[G^n], A)$ where we regard $\mathbb{Z}[G^n]$ as an $H$-module by restricting the $G$-module structure. So we are trying to describe a cochain complex map $\hom_{\mathbb{Z}H}(\mathbb{Z}[H^n], A) \to \hom_{\mathbb{Z}H}(\mathbb{Z}[G^n], A)$, induced by a map from one free $H$-resolution of $\mathbb{Z}$,

$$\ldots \to \mathbb{Z}[G^3] \to \mathbb{Z}[G^2] \to \mathbb{Z}G \to \mathbb{Z} \to 0, \qquad (2)$$

back to the standard one $(1)$. [We know by the standard acyclic models technique that such a map between resolutions induces a canonical isomorphism in cohomology: any two maps such are homotopic.]

Sometimes a map between projective resolutions comes for free. For example, getting a map from (1) to (2) is easy; it is induced by the inclusion $H \hookrightarrow G$. This induces the obvious map you pointed to. But more typically the construction of a map between projective resolutions (in this case of $\mathbb{Z}$) involves a certain amount of choice. For example, to exhibit $\mathbb{Z}[G]$ as a projective (left) $H$-module involves choosing coset representatives, or choosing a splitting $s: H \backslash G \to G$ of the canonical projection $G \to H \backslash G$, so that we can write $\mathbb{Z}[G] \cong \bigoplus_{x \in G/H} \mathbb{Z}[Hs(x)]$.

This general phenomenon occurs frequently, where one has an isomorphism between quotient objects $X/A \to Y/B$ induced by a more or less canonical map $X \to Y$, but the inverse map $Y/B \to X/A$ doesn't come from a canonical $Y \to X$ and one must make choices to extract such a map. A similar example of this phenomenon came up here: "A gentleman never chooses a basis.", where there is in fact a canonical morphism $k \to V \otimes V^\ast$ for $V$ a finite-dimensional space over a field $k$. An explicit set-theoretic description of this map involves choosing a basis, although the map turns out to be independent of basis. It is relevant for that example that a tensor product is inevitably described in terms of generators and relations, i.e., by a quotient construction.

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  • $\begingroup$ If one chooses such a splitting, how would you make this choice into an inverse for the Shapiro Lemma? $\endgroup$
    – Rene Recktenwald
    Commented May 11, 2018 at 11:14
  • $\begingroup$ @Roundthecorner At the $n$-cocycle level, this involves a composite of $H$-module maps $Z^n(H, A) \to Z^n(G, A) \to Z^n(G, \hom_{\mathbb{Z}H}(\mathbb{Z}G, A))$, which is a restriction of a composite of $n$-cochain maps $C^n(H, A) \to C^n(G, A) \to C^n(G, \hom_{\mathbb{Z}H}(\mathbb{Z}G, A))$. The second of these is the canonical isomorphism $\hom_{\mathbb{Z}H}(\mathbb{Z}[G^n], A) \cong \hom_{\mathbb{Z}G}(\mathbb{Z}[G^n], \hom_{\mathbb{Z}H}(\mathbb{Z}G, A))$ mentioned in my answer; it takes $\phi: G^n \to A$ to the map $\psi: G^n \to \hom_{\mathbb{Z}H}(\mathbb{Z}G, A))$ defined by (continued) $\endgroup$
    – Todd Trimble
    Commented May 11, 2018 at 17:28
  • $\begingroup$ $\psi(g_1, \ldots, g_n)(g) = \phi(gg_1, \ldots, gg_n)$. [The inverse of this takes $\psi$ to the map $\phi$ defined by $\phi(g_1, \ldots, g_n) = \psi(g_1, \ldots, g_n)(1)$. If you want to check carefully that this works, remember that "homogeneous" refers to the diagonal action, and that the left $G$-module structure on $\hom_{\mathbb{Z}H}(\mathbb{Z}G, A)$ is defined by $(g \cdot f)(g') = f(g'g)$.] As for the first map $C^n(H, A) \to C^n(G, A)$: that's induced by a map between projective resolutions of the $H$-module $\mathbb{Z}$. A basic result in any text on homological algebra is (cont.) $\endgroup$
    – Todd Trimble
    Commented May 11, 2018 at 17:33
  • $\begingroup$ that there exists a map between any two projective resolutions that is unique up to homotopy (that's what I was referring to in "acyclic models"), and this induces an isomorphism in homology. However, the relevant map $\mathbb{Z}[G^n] \to \mathbb{Z}[H^n]$ between projective resolutions involves choosing an $H$-basis of the free modules $\mathbb{Z}[G^n]$, and that's what I was referring to at the end. $\endgroup$
    – Todd Trimble
    Commented May 11, 2018 at 17:37
  • $\begingroup$ Thank you for the answer. Is it possible to write down explicitely where an element of $H^n(H,CoInd_H^G(A))$ gets mapped to? $\endgroup$
    – Rene Recktenwald
    Commented May 14, 2018 at 7:18

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