In the problem Homotopy relative to closed set, I asked whether 2 maps $f,g:M\to N$ from $M$ to $N$ where $N$ is contractible, such that $f=g$ on some closed set $A\subset M$ are homotopic relative to $A$.
@Paul Frost gives a counterexample with $M=N=$ the comb space. Actually, when I asked the previous question, I was having smooth manifolds and smooth maps in mind. So now I want to know whether this is true or not? That is to say,
Let $f,g:M\to N$ be two smooth maps between smooth manifolds $M$, $N$ where $N$ is contractible. Suppose $f=g$ on a closed set $A\subset M$ , does it follow that $f\simeq g\ {\rm rel}\ A$?
Yes. Every manifold is an ANR (= absolute neighborhood retract for metrizable spaces, see for example https://en.wikipedia.org/wiki/Retract). Each contractible ANR is an AR (= absolute retract for metrizable spaces). So let us prove a more general result:
Let $f,g : M\to N$ be two maps from a metrizable space $M$ to an AR $N$ such that $f=g$ on some closed set $A\subset M$. Then $f,g$ are homotopic relative to $A$.
The proof is very simple. The set $B = M \times \{0,1\} \cup A \times I$ is closed in $M \times I$. Define $\phi : B \to N$ by $\phi(a,t) = f(a)$, $\phi(m,0) = f(m)$ and $\phi(m,1) = g(m)$. This is a contiuous map. Since $N$ is an AR, it has a continuous Extension $H : M \times I \to N$. This is the desired homotopy.
What I am not sure is whether you can always get a smooth homotopy.