# Geometric interpretation of transfer map on homology

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).

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

• Hi igorf, thanks a lot! Does this by any chance also work when $M$ and $N$ are manifolds with boundaries and the $f^{-1}(\partial N)=\partial M$?
– Hans
May 14, 2023 at 11:44
• (see the edits for the precise setup/question)
– Hans
May 14, 2023 at 11:51
• 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. May 14, 2023 at 12:55
• Ah, right. Thanks!
– Hans
May 14, 2023 at 15:48