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

  • 2
    $\begingroup$ 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. $\endgroup$
    – Thomas Rot
    Dec 10, 2023 at 13:30
  • 1
    $\begingroup$ It's true provided $N$ is a sphere, via a framed cobordism argument. $\endgroup$ Dec 11, 2023 at 17:20
  • 2
    $\begingroup$ 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. $\endgroup$ Dec 11, 2023 at 18:40
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 11, 2023 at 20:26
  • 4
    $\begingroup$ 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. $\endgroup$
    – Ian Agol
    Dec 12, 2023 at 0:27

1 Answer 1


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, $\partial B^i\times B^{n-i} \subset H_0\cup \cdots \cup H_{i-1}=H_{<i}$. 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<i} H_{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<i} H_{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 $x\in \cup_{i<n} H_{i-}$).

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 $H_{<i}$ near $\partial B^i\times B^{n-i}$ the map is the identity in the $B^{n-i}$ direction, and in $B^i\times B^{n-i}-H_{<i-}$ the map is $f$.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.