# A question about degree of map on a smooth manifold

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

• 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. Nov 21, 2021 at 10:10
• 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 Nov 21, 2021 at 10:13
• 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. Nov 21, 2021 at 10:36
• @FriederJäckel t is also a regular value Nov 21, 2021 at 10:39
• Could you please write down your details? Thanks again.@FriederJäckel Nov 21, 2021 at 10:50

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 $$$$\partial W=X_a \coprod -X_t,$$$$ where $$-X_t$$ is the manifold $$X_t$$ but with the opposite orientation. So $$$$\mathrm{deg}(\partial g)=\mathrm{deg}(g_a)-\mathrm{deg}(g_t),$$$$ where $$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)$$.

• 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. Nov 21, 2021 at 11:18
• What is your definition of homologous? Or how do you think about it? Nov 21, 2021 at 11:32
• 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 Nov 21, 2021 at 11:38
• 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. Nov 21, 2021 at 11:50
• This should be explained in every introductory book on Algebraic Topology when talking about singular homology, e.g. Chapter 2 in Hatcher‘s book. Nov 21, 2021 at 11:52