Suppose I have a contractible neighborhood $B$ inside a manifold $M$. Then I would like to justify that there exists a family of smooth maps $F_t:M\rightarrow M$ such that $F_0=id$ and $F_1(B)=\{p\}$, where $p\in M$.
We know that $i:B\rightarrow M$ is homotopic to a constant map but other than this I am bit lost on how to get the homotopy,since I am not sure I can use the homotopy extension property , and then how to make it smooth.
Any help is appreciated. Thanks in advance.
As requested, I’m turning my hint into a fully-fledged solution.
First, as Connor Malin’s comment points out, the statement is false in general, because some manifolds can have a dense contractible open subset. A general example would be any compact manifold which has a finite cellular decomposition with a single top-dimensional cell (the interior of which is open, dense and contractible) – real and complex projective spaces, spheres, tori are in this case.
A natural restriction (not the only one) is if we require $B$ to be small enough. If $B$ is small enough, we can assume that there is a smooth, open, injective map $\pi: 3B^n \rightarrow M$ which is a diffeomorphism into its image $B’$ and such that $\pi(B^n) \supset B$.
Let $\psi:[0,3] \rightarrow [0,1]$ as a smooth function such that $\psi([0,1])=\{0\},\psi([2,3])=\{1\}$. Let, for $0 \leq t \leq 1$, $g_t: x \in 3B^n \longmapsto (1-t+t\psi(\|x\|))x \in 3B^n$. Then if $\|x\| \geq 2$, $f_t(x)=x$; $f_0(x)=x$ for any $x$; and $f_1(x)=0$ if $\|x\|=1$.
Define, for $q \in M$, $f_t(q)=q$ if $q \notin B’$, and $f_t(q)=\pi(g_t(\pi^{-1}(q)))$ if $q \in B’$. Then $f_t$ is a smooth sequene of smooth maps $M \rightarrow M$; $f_0=id$ and $f_1(B)=\{\pi(0)\}$.