How is this space a deformation retract and not a strong deformation retract?

Question

From a paragraph is Rotman's Algebraic Topology, could someone explain how this is a deformation retract and not a strong deformation retract?

What would be the deformation retract for this space?

Answer 1

For the first part, let $F_t : [0, 1] \times X \to X$ be a transformation given by the formula $F_t(p) = (1-t) \cdot p$. It is continuous, $F_0 = \operatorname{id}_X$ and $F_1(p) = 0$ for all $p \in X$.

Let $G_t : [0, 1] \times X \to X$ be a transformation given by $G_t( p ) = t \cdot \pi_x(p)$ where $\pi_x : X \to I$ is the orthogonal projection. Then $G_t$ is continuous, $G_0(p) = 0$ for all $p \in X$ and $G_1[X] = I$.

Clearly the concatenation of these two transformations is a deformation retraction of $X$ onto $I$.

[Credit for the nice word 'concatenation' goes to Lee Mosher. ;p]

Intuitively, we pull everything to $0$ in the first phase and expand it to $I$ in the second phase.

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For the second part, suppose $H : [0, 1] \times X \to X$ is a strong deformation retraction onto $I$. Since

$$V = \left\{ (x, y) \in X : x > \frac{1}{2} \right\}$$

is an open neighbourhood of the point $\left< 1, 0 \right>$ in $X$, the preimage $G = H^{-1}[V]$ is open. Moreover $[0, 1] \times \{ \left< 1, 0 \right> \} \subseteq G$ because $H$ is a strong deformation. By compactness of $[0, 1]$, there is open neighbourhood $U \subseteq X$ of $\left< 1, 0 \right>$ such that $[0, 1] \times U \subseteq G$.

Let $p \in U \setminus I$, specifically $p \in U \cap I_n$ for some $n \in \mathbb{N}$, where $I_n$ is the segment with slope $\frac{1}{n}$. Now $H \big[ [0, 1] \times \{ p \} \big]$ is a connected subset of $X$, and also a subset of $V$, so it must be a subset of the connected component of $H( 0, p ) = p$ in $V$ which is $V \cap I_n$. So $H(1, p) \in V \cap I_n$, thus $H(1, p) \notin I$ which is a contradiction.

Intuitively, if $I$ is fixed by $H$ throughout time, a small neighbourhood of $\left< 1, 0 \right>$ must stay close to $\left< 1, 0 \right>$ during the transformation, but it can't end up being moved to $I$ without going through the origin, which is far from $\left< 1, 0 \right>$.