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Let $f\colon M\to N$ a smooth surjective map of compact oriented manifolds of the same dimension. Then there is a map $f_!\colon H_i(N)\to H_i(M)$ obtained from the induced map on cohomology combined with Poincaré duality. This map has several names. I have seen it called the transfer, umkehr or wrong-way map. And it is treated in several question on mathoverflow and math.stackexchange:

Pullback map in homology

Reference for push-pull formula in cohomology

https://math.stackexchange.com/questions/4683875/pullback-of-homology-via-trace-and-poincar%C3%A9-duality

I am interested in a geometric interpretation of this map in the following situation. Assume that there is an open subset $U\subset N$ such that, letting $V=f^{-1}(U)$, the restriction $f|_V\colon V\to U$ is an $r$-sheeted covering map ($r\in\mathbb{N}$) and $N'\subset U$ a compact oriented submanifold. If I understood correctly the answers and comments from the questions above, then we have

$f_!([N'])=[f^{-1}(N')]$.

Is this correct? And is there a reference for this? Several books in the answers and comments to the questions above are mentioned but I have actually not found this statement.

I should also say that I want to use this statement in a paper (if true!) and neither me nor the audience of the paper are experts on algebraic topology. So I would very much appreciate a reference where this fact is explicitly stated or at least a statement which easily implies this fact.

Edit: One can construct $f_!$ also in the case when $M$ and $N$ are oriented manifolds with boundary and $f^{-1}(\partial N)=\partial M$ (and all other assumptions the same). Now the map map $f_!\colon H_i(N)\to H_i(M)$ is obtained from the induced map on cohomology relative to the boundaries combined with Lefschetz duality. Is the conclusion $f_!([N'])=[f^{-1}(N')]$ still true? We still assume $f|_V\colon V\to U$ being a $r$-sheeted covering, neither $U$ nor $V$ intersect the boundaries of $M$ and $N$, and $N'\subset U$ a compact oriented submanifold (without boundary).

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Let $K\subset N$ be a compact oriented smooth submanifold of codimension $k$ in an oriented smooth manifold $N$, $T\subset N$ a tubular neighbourhood of $K$ and $\tau$ a $k$-form on $N$ supported on $T$ such that the restriction of $\tau$ to the fibre of $T$ over any point of $K$ (which is of course diffeomorphic to $\mathbb R^k$) has compact support and integrates to 1. Then the class of $\tau$ in $H^k_c(N)$ is Poincaré dual to $[K]$. Such a $\tau$ always exists; it is called a Thom form. This is discussed in Bott, Tu, Differential forms in algebraic topology, § 5--6.

If $M$ is an oriented smooth manifold, $f:M\to N$ a smooth map transversal to $K$ and $T$ a sufficiently small tubular neighbourhood of $K$, then $f^{-1}(T)$ is a tubular neighbourhood of $f^{-1}(K)$; one can arrange trivializations in such a way that $f$ maps any fibre of $f^{-1}(T)$ diffeomorphically to a fibre of $T$. So the pullback via $f$ of a Thom form for $K$ will be a Thom form for $f^{-1}(K)$. This proves the desired statement $f_!([K])=[f^{-1}(K)]$.

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  • $\begingroup$ Hi igorf, thanks a lot! Does this by any chance also work when $M$ and $N$ are manifolds with boundaries and the $f^{-1}(\partial N)=\partial M$? $\endgroup$
    – Hans
    May 14, 2023 at 11:44
  • $\begingroup$ (see the edits for the precise setup/question) $\endgroup$
    – Hans
    May 14, 2023 at 11:51
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    $\begingroup$ Hi Hans! Let's remove the boundaries. This doesn't change homology, and the map $\overset\circ f:\overset\circ M\to\overset\circ N$ remains proper. Is your $f_!$ equal to the composition $H_*(N)\simeq H_*(\overset\circ N)\simeq H^*_c(\overset\circ N)\xrightarrow{f^*} H^*_c(\overset\circ M)\simeq H_*(\overset\circ M)\simeq H_*(M)$? If so, then the same reasoning works just as well. $\endgroup$
    – igorf
    May 14, 2023 at 12:55
  • $\begingroup$ Ah, right. Thanks! $\endgroup$
    – Hans
    May 14, 2023 at 15:48

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