Deformation retraction restricted to a path

Question

Let $X$ be a topological space with subspace $Y$. Suppose that we have a (strong) deformation retraction onto $Y$. That is, a continuous map $F:X\times [0,1]\to X$ such that for all $x\in X$, $F(x,0)=x$, $F(x,1)\in Y$, and for all $y\in Y$ and all $t\in [0,1]$, $F(y,t)=y$.

Is it true that for any $x\in X$ and $t\in [0,1]$, $F(F(x,t),1)=F(x,1)$? It corresponds to my intuition regarding deformation retracts as paths $F(x,\cdot):I\to X$ in $X$ such that all points on the path have the same endpoint in $Y$. However I was unable to prove the result. Any thoughts on this? Any help is much appreciated.

Answer 1

No. Let $B_r$ denote the closed ball of radius $R$ on $\mathbb{C}$ (around the origin), and take $X=B_2$, $Y=B_1$.

Think of $F$ as "rotating and shrinking". More precisely, define the "shrinking part" (at time $t$) as $$s_t(x)=\begin{cases} 1&\text{, if }0\leq x\leq 1\\ \left(1-\frac{t}{2}(x-1)\right)&\text{, if }1\leq x\leq 2 \end{cases}$$ and the "rotating part" as $$r_t(x)=e^{2\pi i t}\cdot x$$ (where "$\cdot$" denotes complex multiplication), and then consider $$F(x,t)=s_t(\Vert x\Vert)r_t(x)$$

Then $F(F(x,t),1)=F(x,1)$ if and only if $t=0$ or $1$.

Answer 2

The claim is not true. Consider $G$ a strong deformation of the square $[-1,1]\times [-1,1]$ onto the segment $\{0\}\times [-1,1]$ such that $G((1,1),1)\neq (0,1)$ - for instance, a deformation that squashes diagonally the square onto the mid segment.

Now, consider the deformation $F((x,y),t)=\begin{cases}((1-4t)x,y)&\text{if }t\le\frac12\\ G((-x,y),2t-1)&\text{if }t>\frac12\end{cases}$

Since $F((x,y),1/2)=(-x,y)=G\left((-x,y),2\frac12-1\right)$ and $F$ leaves fixed the points with $x=0$, this is actually a deformation of $[-1,1]\times [-1,1]$ onto $\{0\}\times [-1,1]$

However, $F((-1,1),1)=G((1,1),1)\ne (0,1)$, while $$F\left(F\left((-1,1),\frac14\right),1\right)=F((0,1),1)=(0,1)$$

Added: More in general, if your claim were true, then the map $x\mapsto F(x,1)$ would be independent from which strong deformation of $X$ onto $Y$ you are considering. In fact, for two strong deformations $F_1$ and $F_2$, you could consider $$F(x,t)=\begin{cases} F_1(x,3t)&\text{if }t\le\frac13\\ F_1(x,2-3t)&\text{if }\frac13<t\le\frac23\\ F_2(x,3t-2)&\text{if }t>\frac23\end{cases}$$

And use the "identities" $$\begin{cases}F(x,1)\color{red}{=F(F(x,1/3),1)}=F_2(F_1(x,1),1)=F_1(x,1)\\ F(x,1)=F_2(x,1)\end{cases}$$

However, there is no such result!

Answer 3

Thank you for your answers. Indeed I now realize that the conjecture is false. In fact, I came up with a quite a silly (but easy) counterexample myself:

Let $X=\{a,b,c\}$ with trivial topology, $Y=\{a,b\}$. Define $F:X\times [0,1]\to X$ (continuous because $X$ has the trivial topology) by $$ F(a,t)=a,\quad F(b,t)=b \quad \mbox{ for all } t\in[0,1],\quad F(c,t)=\begin{cases} c &\mbox{ if }\quad t=0\\ a &\mbox{ if }\quad t\in(0,1)\\ b &\mbox{ if }\quad t=1 \end{cases} $$

Now (similarly to the example by Luiz) $$ F(F(c,t),1)=F(c,1)\iff t\in \{0,1\} $$