Homotopy equivalent Morse functions 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?
 A: 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 $t>0$.
