I'm a little confused about the following:

Let $X,Y$ be two smooth manifold, $f: X\to \mathbb{R}^+$ be a smooth map on $X$. Let $b>a>0$ be regular values of $f$. If we consider the compact manifold

$$X_t:=f^{-1}(t),\qquad \text{here $t$ is any regular value of $f$} $$

then my question is as follows:

(1)Why is $X_t$ homologous to $X_a:=f^{-1}(a)$?

(2)Additionally, if there is a map $X\to Y$, then why does the degree of maps $$ X_t\to Y$$ $$ X_a\to Y$$ from $X_t$ and $X_a$ to smooth manifold $Y$ respectively must coincide.

Since it is not obvious to me, could you please give me some help with more details? Thanks

  • $\begingroup$ If you don‘t assume any further conditions on the maps $X_t \to Y$ and $X_a \to Y$ it is impossible to conclude that they have the same degree. If you assume that there is a map $X \to Y$, then this is indeed true. $\endgroup$ Nov 21, 2021 at 10:10
  • $\begingroup$ Yes, under the assumption. Could you give me some details about the solution to my two questions? Because it is not obvious to me. Thanks $\endgroup$ Nov 21, 2021 at 10:13
  • $\begingroup$ I can give you an answer when $t$ is also a regular value of $f$. I think this is no restriction (at least for the second question) since when talking about degree you need to know that $X_t$ is a manifold, which in general is only guaranteed if $t$ is a regular value. $\endgroup$ Nov 21, 2021 at 10:36
  • $\begingroup$ @FriederJäckel t is also a regular value $\endgroup$ Nov 21, 2021 at 10:39
  • $\begingroup$ Could you please write down your details? Thanks again.@FriederJäckel $\endgroup$ Nov 21, 2021 at 10:50

1 Answer 1


I give my answer under the assumption that $t$ is also a regular value of $f$.

Since $a$ and $t$ a regular values for $f$ the preimage $W:=f^{-1}([a,t])$ is a bordism from $X_a$ to $X_t$, that is $W$ is a smooth manifold of dimension $\mathrm{dim}(X)$ with boundary $\partial W=X_a \coprod X_t$.

This already answer your question (1) since the fact that $X_a$ and $X_t$ are bordant implies that they are homologous. You can see this by choosing a triangulation of $W$.

For question (2) we need the following fact (Theorem 17.38 in Lee‘s Introduction to Smooth Manifolds $2^{nd}$ edition): For a restriction $\partial g: \partial W \to Y$ of a map $g: W \to Y$ it holds $\mathrm{deg}(\partial g)=0.$

Recall that the choice of orientation is important for the value of the degree (changing the orientation changes the sign of the degree). Note that \begin{equation} \partial W=X_a \coprod -X_t, \end{equation} where $-X_t$ is the manifold $X_t$ but with the opposite orientation. So \begin{equation} \mathrm{deg}(\partial g)=\mathrm{deg}(g_a)-\mathrm{deg}(g_t), \end{equation} where $g_a:X_a \to Y$ and $g_t:X_t \to Y$ are the restrictions of $g: X \to Y$. This implies $\mathrm{deg}(g_a)=\mathrm{deg}(g_t)$.

  • $\begingroup$ I'm a little confused about choosing a triangulation of $W$ to prove that $X_a$ and $X_t$ are homologous if $X_a$ and $X_t$ are cobordant. I don't know how to finish your idea in detail. Could you please explain it more? Thanks in advance. $\endgroup$ Nov 21, 2021 at 11:18
  • $\begingroup$ What is your definition of homologous? Or how do you think about it? $\endgroup$ Nov 21, 2021 at 11:32
  • $\begingroup$ I'm actually unfamiliar with the definition of homologous because it just appears in a proof of a theorem of an article of Gromov. Could you please explain its usual definition to me? Thanks $\endgroup$ Nov 21, 2021 at 11:38
  • $\begingroup$ It states ''homologous'' to imply the degree of the above two maps coincides in Gromov's article. As your answer, I think the statement of ''homologous'' seems to be redundant. $\endgroup$ Nov 21, 2021 at 11:50
  • $\begingroup$ This should be explained in every introductory book on Algebraic Topology when talking about singular homology, e.g. Chapter 2 in Hatcher‘s book. $\endgroup$ Nov 21, 2021 at 11:52

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