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



When is finding an explicit inverse of an isomorphism not possible
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."](https://mathoverflow.net/questions/400/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. 


