So I learn about these two things today. Let $M,N$ be smooth manifolds and $f_1,f_2$ be smooth maps from $M$ to $N$
Def: $f_1$ and $f_2$ are HOMOTOPIC if there exists a a smooth map $H: [0,1]\times M\to N$ s.t. $H(0,m)=f_1(m)$ and $H(1,m)=f_2(m)$.
Def: $f_1$ and $f_2$ are SMOOTHLY HOMOTOPIC if there exists a a smooth map $H: \mathbb{R} \times M\to N$ s.t. $H(0,m)=f_1(m)$ and $H(1,m)=f_2(m)$.
I want to ask what is the relation between these two. I understand that if they are smoothly homotopic, then they are homotopic since we can restrict the homotopy map. What about the other way around? If we have a homotopy on $[0,1]$, can we extend it to $\mathbb{R}$?
Suppose that $H:[0,1]\times M\rightarrow N$ is a smooth homotopy. We are working here with a manifold with boundary, so what does it mean that this map is smooth?
Well, we have to recall, what does it mean that a map $f:H^+\rightarrow N$ is smooth, where $H^+\subseteq \mathbb{R}^n$ is a halfspace. One of the definitions (that I am aware of) is that there exists an open subset $H^+\subseteq U\subseteq \mathbb{R}^n$ such that $f$ extends to a smooth map $\tilde{f}:U\rightarrow N$. If we apply this definition to a homotopy $H:[0,1]\times M\rightarrow N$ and use compactness of $[0,1]$, we derive that there exists open subset $[0,1]\subseteq U\subseteq \mathbb{R}$ such that $H$ can be extended to a smooth map $\tilde{H}:U\times M\rightarrow N$. We may assume that $U$ is an open interval containing $[0,1]$. Next we can pick a diffeomorphism $d:\mathbb{R}\rightarrow U$ such that $d_{\mid [0,1]}= 1_{[0,1]}$ and finally we define $$F = \tilde{H}\cdot (d\times 1_M):\mathbb{R}\times M\rightarrow N$$ This is a homotopy in the other sense extending $H$.