Suppose $X$ is a (smooth) manifold with boundary and $f: [0,\infty)\times \partial X \rightarrow X$ is an open embedding such that $f(0,x) = x (\forall x \in \partial X) $ i.e., a collar. Is $f([0,1] \times \partial X)$ closed in $X$ ?
It seems to be true but I have some problems: Suppose $f(t_n,x_n) \to y\in X$, where $t_n \in [0,1]$ and $x_n \in \partial X$. I try to show that $y = f(t,x)$ for some $t\in [0,1]$ and $x\in \partial X$. But I don't know how to deal with the case where $y \in X-Im(f)$. Thank you.
Here is an example.
Let $X$ be the upper half-plane $\{(x,y): y\ge 0\}$. Consider the map $$ f: (x,t) \mapsto ((1-t)x + t(e^x+1), t), \quad 0\le t\le 1, x\in {\mathbb R}. $$ This map restricts to the identity map when $t=0$ and sends ${\mathbb R}\times [0,1]$ onto $$ U={\mathbb R}\times [0,1) \cup (0,\infty)\times \{1\}\subset X, $$ which is the strip $0\le y\le 1$ with a ray removed from the top. Hence, $U$ is not closed. I am leaving it to you to check that $$ f: {\mathbb R}\times [0,1]\to U$$ is a diffeomorphism.