There are many discussions has done on Hatcher 0.5, some here, and some there. But I am primarily uncertain about the invoke of tube lemma. Wonder if someone would kind help me take a look at my proof and help me verify or revise? Thank you.
Show that if a space $X$ deformation retracts to a point $x \in X$, then for each neighborhood $U$ of $x$ in $X$ there exists a neighborhood $V \subset U$ of $x$ such that the inclusion map $V\subset U$ is nullhomotopic.
Given $X$ deformation retracts to a point $x \in X$, we know there is a family of maps $f_t: X \to X, t \in I$, such that $f_0 = \mathbb{I}$ (the identity map), $f_1(X) = x$, and $f_t|x= \mathbb{I}$ for all $t$.
Tube Lemma: Let $X$ and $Y$ be topological spaces with $Y$ compact, and consider the product space $X \times Y$. If $N$ is an open set containing a slice in $X \times Y$, then there exists a tube in $X \times Y$ containing this slice and contained in $N$.
To make sure $f_t(V) \subset V$, we need to find the appropriate $V$. By the definition of retraction, we know that $f_t(x) = x$, that is the slice $x \times I$ is contained in some open set $N$, and hence by tube lemma, $N$ also contains a tube containing the slice $x \times I$. Hence, let $V = N$.
Now we consider the inclusion map $i: V \to U$, and consider another family of maps: $\tilde{f}_t = f_t \circ i: X \to X, t \in I$ such that $\tilde{f}_0 = \mathbb{I} \circ i = i, \tilde{f}_1 = c \circ i$, $\tilde{f}_t|c= \mathbb{I}$ for all $t$.
Hence we constructed a homotopy map from inclusion map to a constant map, hence it is nullhomotopic.
