# Smooth map between oriented manifolds

Let $$f: M\rightarrow N$$ be a smooth map between smooth closed oriented connected manifolds of same dimension.

Question: is it true that $$f$$ is smoothly homotopic to some smooth map $$g: M\rightarrow N$$ such that

• The determinant of the differential $$\det (dg(m))\geq 0$$ for all $$m\in M$$
• or $$\det(dg(m))\leq 0$$ for all $$m\in M$$.

PS: this is a duplicate, I asked this question here: https://math.stackexchange.com/questions/4812807/smooth-map-between-oriented-manifolds?noredirect=1#comment10258150_4812807

• If the map is not surjective it is homotopic to a map with differential zero. Namely you can homotope the map to have image in the n-1 skeleton. Dec 10, 2023 at 13:30
• It's true provided $N$ is a sphere, via a framed cobordism argument. Dec 11, 2023 at 17:20
• Presumably the homotopy-equivalences of non-diffeomorphic lens spaces would be the most elementary counter-examples. So the homotopy-equivalence $L_{7,1} \to L_{7,2}$ would presumably be what you are looking for. I haven't worked out a complete argument but I think that should work. Dec 11, 2023 at 18:40
• If I have time I'll try to flesh out the argument. But I think a computation of the Whitehead torsion of the map should likely do the job. Dec 11, 2023 at 20:26
• I think this should follow from a result of Hopf (see doi.org/10.1112/plms/s3-16.1.369) and the following: Take a handlebody decomposition of N (with one n-handle), then there is a smooth map $N\to N$ so that each handle is retracted to its core. I haven’t checked this carefully, but it might be true. en.wikipedia.org/wiki/Handle_decomposition Then compose the map given by Epstein with this pinch map to get a map with the desired property. Dec 12, 2023 at 0:27

This follows from a result of Hopf (see the exposition of Epstein). By this result, one may assume that there is a disk $$D\subset N$$ such that $$f^{-1}(D)$$ is $$d$$ disks mapping diffeomorphically to $$D$$, where $$d$$ is the degree of the map (we may assume $$d$$ is positive by switching orientation if needed).
Claim: Let $$D\subset N$$ be a ball. Then there is a map (homotopic to the identity) $$h: N\to N$$ such that $$det(dh_x)=0$$ for all $$x\in N-D$$ and $$det(h_x)\geq 0$$ for all $$x$$.
To prove this, cover $$N$$ with a smooth handle decomposition. This is a covering of $$N$$ by closed neighborhoods $$H_0, \ldots, H_n$$, $$n=dim N$$ (we assume the interiors cover $$N$$). Each $$H_i$$ is smoothly diffeomorphic to a disjoint union of $$i$$-handles, that is balls with a diffeomorphism to $$B^i \times B^{n-i}$$. Moreover, for each $$i$$-handle, $$\partial B^i\times B^{n-i} \subset H_0\cup \cdots \cup H_{i-1}=H_{. The existence of a smooth handle decomposition follows eg from Morse theory. Moreover, we may assume that $$H_n$$ is a single $$n$$-handle which we may identify with our ball $$D$$ by a diffeomorphism.
Choose slightly smaller handle neighborhoods $$H_{i-}$$ with handles $$B^i\times B_-^{n-i} \subset B^i\times B^{n-i}$$ so that the interiors of these handles cover as well. We have smooth maps $$g_i: N\to N$$ so that $${g_i}_{|N-H_i}=Id$$, $$g_i(H_i)=H_i$$, $$det((dg_i)_x)=0$$ for all $$x\in B^i\times B_-^{n-i} - \cup_{j, $${g_i}_{|B^i\times B^{n-i}}$$ preserves the $$\{x\}\times B^{n-i}$$ slices on the handle, $$g_i$$ sends $$B^i \times B_-^{n-i}$$ to $$B^i\times \{0\}$$ outside of $$\cup_{j , and $$det((dg_i)_x)\geq 0$$ for all $$x\in N$$ *. Then the composition $$h=g_{n-1}\circ \cdots \circ g_0$$ will have $$det(dh_x)\geq 0$$ for all $$x\in N$$ and $$det(dh_x)=0$$ for $$x\in N-D$$ (since by construction $$det(dh_x)=0$$ for $$x\in \cup_{i).
Then $$h\circ f$$ will have the desired property.
• The map $$g_i$$ restricted to each handle preserves the $$\{x\}\times B^{n-i}$$ directions on each handle. In the $$B^{n-i}$$ direction one uses a bump function (such as my favorite the Fabius function) to make a function $$f:B^{n-i}\to B^{n-i}$$ sending $$B^{n-i}_-$$ to $$0$$ (see the diagram for a picture in the radial coordinate). Then use a partition of unity in the $$B^i$$ radial coordinate so that in $$H_{ near $$\partial B^i\times B^{n-i}$$ the map is the identity in the $$B^{n-i}$$ direction, and in $$B^i\times B^{n-i}-H_{ the map is $$f$$.