Does strong deformation retraction move points along fixed x curve along that curve?

Question

Let $X$ be a topological space, and $f : X \times [0, 1] \to X$ be a strong deformation retraction from $X$ to $X' \subset X$. Does it hold that for each $x \in X$ and $t \in [0, 1]$,

$$f(f(x, t), 1) = f(x, 1)$$

?

This property does not hold for deformation retractions in general. For example, let $X = X' = [0, 1]$, and $f(x, t) = (1 - t + t^2)x$. Then $f$ is continuous, $f(x, 0) = x$, and $f(x, 1) = x \in X'$; i.e. $f$ is a deformation retraction from $[0, 1]$ to itself. However, $f(f(1/2, 1/2), 1) = f(1/2, 1/2) = 3/8$ while $f(1/2, 1) = 1/2$.

A strong deformation retraction from $X$ to $X' \subset X$ is a continuous function $f : X \times [0, 1] \to X$ such that for each $x \in X$, $x' \in X'$ and $t \in [0, 1]$,

$$f(x, 0) = x,$$ $$f(x, 1) \in X',$$ $$f(x', t) = x'.$$

Answer 1

No, this is not always the case: The following example is rather synthetic, but I'm sure there are examples in $\mathbb{R}^n$.

Consider $\mathbb{R}$ with the indiscrete topology. Then any $f: [0,1] \times \mathbb{R} \rightarrow \mathbb{R}$ is continuous. We want to deformation retract $[0,1] \hookrightarrow \mathbb{R}$. Then define for $t \in [0,\frac{1}{2}]$:

$$ \begin{cases} f(x,t) = 1, & x < 0 \\ f(x,t) = x, & x \in [0,1] \\ f(x,t) = 1, & x > 1 \end{cases} $$

and for $t \in (\frac{1}{2},1]$:

$$ \begin{cases} f(x,t) = 0, & x < 0 \\ f(x,t) = x, & x \in [0,1] \\ f(x,t) = 1, & x > 1 \end{cases} $$

Then we have a strong deformation retract since $\forall t \in [0,1], x \in [0,1]: f(x,t) = x$, but for $t = \frac{1}{3}$ and $x = -1$ we have $f(f(-1,\frac{1}{3}),1) = f(1,1) = 1 \neq 0 = f(-1,1)$.

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A more illustrating example is: $[0,1] \hookrightarrow \mathbb{R}^2$, with the homotopy given as first applying the matrix $\begin{pmatrix} 1, t \\ 0, 1 \end{pmatrix}$ for $t \in [0,\frac{1}{2}]$ and then (again smoothly) mapping $(1,1) \mapsto (0,0)$ and $(0,1) \mapsto (0,1)$ and extending linearly.

Answer 2

No. Let $\phi : I \to I, \phi(t) = 2t$ for $t \le 1/2$, $\phi(t) = 1 -2t$ for $t \ge 1/2$.

Let $X = \mathbb R \times I$ and $f(x,y,t) = (x + \phi(t)y,(1-t)y)$. Then $f(x,y,0) = (x,y), f(x,y,1) =(x,0)$ and $f(x,0,t) = (x,0)$. That is, $f$ is a strong deformation retraction to $X' = \mathbb R \times \{0\}$. But $$f(f(x,1,t),1) = f(x + \phi(t),1-t,1) = (x+\phi(t),0) \ne (x,0) = f(x,1,1)$$ for $t \in (0,1)$.