my question is the following: given a smooth manifold $M$, take a homotopy of maps $f\_{t}:M \rightarrow \mathbb{R}, \quad t \in [0,1]$ such that every $f\_{t}$ is a Morse function. Do $f\_{0}$ and $f\_{1}$ have the same critical points' structure?
In that case, is it possible to generalize the result to Morse-Bott functions?



Homotopy equivalent Morse functions
To see why the answer is yes for the first question use the implicit function theorem. For any $t\_0$ you can find $\newcommand{\ve}{\varepsilon}$ $\ve >0$ and smooth maps


$$\gamma\_1,\dotsc, \gamma\_N:(t\_0-\ve,t\_0+\ve)\to M$$
 such that for any $t\in (t\_0-\ve,t\_0+\ve)$ the points $\gamma\_1(t),\dotsc, \gamma\_n(t)\in M$ are pairwise distinct and the critical set of $f\_t$ consists precisely of these points. Clearly the index of $\gamma\_i(t)$ is independent of $t$, for any $i$.


If we denote consider the set


$$ C:=\bigl\lbrace (t,x)\in [0,1]\times M;\;\;df\_t(x)=0\bigr\rbrace, $$


then the above fact shows that the natural projection


$$ C\ni (t,x)\mapsto t\in [0,1] $$


is a covering map. 


As for the second question, I will give an example. Denote by $\DeclareMathOperator{\Gr}{Gr}$ by ${\Gr}\_k(V)$ the Grassmannian of $k$-dimensional subspaces of the complex Hermitian space $V$. For any selfadjoint operator $A: V\to V$ we have a smooth function


$${\Gr}\_k(V)\ni L \mapsto f\_A(L)= {\rm Re}\; {\rm tr}(AP\_L)\in \mathbb{R}, $$


where $P\_L$ denotes the orthogonal projection onto $L$. This function is Morse-Bott for any $A$, but it is Morse if and only if the eigenvalues of $A$ are simple, i.e., pairwise distinct. You can now easily construct a smooth family of selfadjoit operators $A\_t$, $t\in [0,1]$ such that $A\_0=\boldsymbol{1}\_V$ but $A\_t$ has simple eigenvalues for any $t>0$.


