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
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 $\mathrm{deg}(g_a)=\mathrm{deg}(g_t)$.