# A deformation retract that is not a strong deformation retract

In Lee's Introduction to Topological Manifolds, problem 7-12 asks to show that $\{(1,0)\}$ is a deformation retract, but not a strong deformation retract of the subspace of the plane

$$X = \bigcup_{n=0}^\infty L_n$$

where $L_0$ is the line segment connecting $(1,0)$ to the origin and $L_n$ is the line segment connecting the origin to $(1,1/n)$ for $n \geq 1$. The part I'm struggling with is to show that it is not a strong deformation retract. If it were, then there would be a continuous map $H: X \times [0,1] \rightarrow X$ with

$$H(x,0) = x, \quad \forall x \in X;\\ H(x,1) = (1,0), \quad \forall x \in X; \\ H((1,0),t) = (1,0) \quad \forall t \in [0,1]$$

I was hoping a contradiction would pop out somewhere, but no luck. Can someone give a hint or suggest a different approach?

• How does one prove that $(1,0)$ is a deformation retract of $X$? – Eduardo Longa Nov 25 '18 at 23:58
• Hi @EduardoLonga. If you haven't figured it out yet, I would suggest making this a separate question. I'm guessing I first found an explicit homotopy that pushed every point to the origin $(0,0)$ along straight lines...and you can guess what the straight lines in $X$ are. Then you just have to push the origin to $(1,0)$ along $L_0$. If you draw a picture of the space, it should be clear that this is the only way to send points in $L_n$ to $(1,0)$ when $n\geq 1$. User12345's answer below shows that continuity requires you do the same to points in $L_0$. – Philip Hoskins Nov 28 '18 at 19:13

OUTLINE: Let $H:X\times[0,1]\to X$ with $H|_{X\times\{0\}}=\mathrm{id}_X$ and $H|_{X\times\{1\}}=c_{(1,0)}$ the constant map to $\{(1,0)\}$ be continuous. Consider the sequence of points $p_n=(1,\frac{1}{n})$. Now prove (or recall) the following facts:

1. There is a characterization of continuity from elementary analysis in terms of convergent sequences in the domain (it holds in general for Hausdorff spaces).

2. Every bounded infinite sequence of reals has a convergent subsequence.

3. The sequence $p_n\to(1,0)$ as $n\to\infty$.

4. For all $n\in\mathbb{Z}^+$, every path $\alpha_n$ from $p_n$ to $(1,0)$ in $X$ has a time $t_n\in[0,1]$ such that $\alpha_n(t_n)=(0,0)$.

5. The map $H|_{\{p_n\}\times[0,1]}$ determines a path from $p_n$ to $(1,0)$ for all $n\in\mathbb{Z}^+$.

Used together these facts will imply that there is a time $t_0\in[0,1]$ such that $H((1,0),t_0)=(0,0)$; in particular $H$ does not fix $(1,0)$ and is thus not a strong deformation retraction map onto $\{(1,0)\}$. As $H$ was arbitrary there can be no strong deformation retraction to $\{(1,0)\}$ as desired.

• This makes sense.Thanks. – Philip Hoskins Jun 28 '16 at 12:56
• Glad I could help! – User12345 Jun 28 '16 at 13:36