Equivalence of two definitions of smoothly homotopic (Setting: $M$,$N$ :smooth manifolds with boundary,  $f_0$,$f_1$: $M\to N$:smooth maps)
There are (at least) two definitions of smoothly homotopic:
$\bf{(1)}$ If there is a smooth map $H:M\times \mathbb{R}\to N$ such that $H(-,i)=f_i\ \ (i=0,1)$, then $f_0$ and $f_1$ are smoothly homotopic.
$\bf{(2)}$ If there is a smooth map $H:U\to N$ where $U$ is some neighborhood of $M\times I$ in $M\times \mathbb{R}$ such that $H(-,i)=f_i\ \ (i=0,1)$, then $f_0$ and $f_1$ are smoothly homotopic.
Tu's book ("An Introduction to Manifolds") adopts (1), and Lee's book ("Introduction to Smooth manifolds") adopts (2). These two definitions are equivalent? If so, could you tell me its proof or any references?
 A: I will prove $\mathbf{(2)}\implies \mathbf{(1)}$. Consider a smooth bump function $\mathcal B\colon \Bbb R\to \Bbb R$ such that
$a)$ $\mathcal B(x)=0$ for $x\leq 0$;
$b)$ $0<\mathcal B(x)<1$ for $0<x<1$;
$c)$ $\mathcal B(x)=1$ for $x\geq 1$.

Let $M, N$ be smooth manifolds, and $H\colon U\to N$ be a smooth function, where $U$ is an open neighborhood of $M\times [0,1]$ in $M\times \Bbb R$.
Define, $\overline H\colon M\times \Bbb R\to N$ as $$\overline H(x,t):=H\big(x,\mathcal B(t)\big)\text{ for all }(x,t)\in M\times \Bbb R.$$ Note that for each $x\in M$ we have $\varepsilon_x>0$ such that $\{x\}\times (0-\varepsilon_x,1+\varepsilon_x)\subseteq U$. So, $\overline H$ is a well-defined smooth function, such that $\overline H(-,0)=H(-,0)$ and $\overline H(-,1)=H(-,1)$.

To construct such $\mathcal B$ consider $\psi:\Bbb R\to \Bbb R$ given by $$\psi(x)=\begin{cases}\exp\left(\frac{1}{x(x-1)}\right)& \text{ if }0<x<1,\\ 0 & \text{ otherwise}.\end{cases}$$ Now, consider $$\mathcal B(x):=\frac{\displaystyle\int_0^x\psi(t)\ \mathrm{d}t}{\displaystyle\int_0^1\psi(t)\ \mathrm{d}t}\text{ for all }x\in \Bbb R.$$
A: Clearly (1) implies (2). The fact that (2) implies (1) appears to be somewhat subtle.
In case that the manifolds have no boundary, this follows from the collar neighborhood theorem (let me call it CNT). In any case, take a collar neighborhood of $U\smallsetminus (M\times [0,1])$. This has two components, each diffeomorphic to $M\times [0,1)$; choose your favorite smooth map $[0,1)\to\Bbb R_{\geq 0}$ and pull it back to the components. Everything in sight is smooth so the appropriate compositions will transform $H$ from a map on $V \cup (M\times [0,1])$ (not $U$!) to a smooth map $M\times [0,1]\to N$.
This result is intuitive but the details of the CNT do require some machinery, so I wonder if there is a simpler argument. To this end, when the manifolds do have boundary, the same argument should work except that you need a suitable CNT variant. Apparently nobody bothers to write such things down, although it seems to me that Lemma 2.1.6 of this paper is equivalent.
(NB: The first answer was completely incorrect, hence the odd first comments.)
