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Not really research level but here goes anyway: Suppose I have a topological space $X$ with a closed subset $K$ and a continuous map $f : X \times [0,1) \to X$ such that:

0) $f(x,0) = x$.

1) For all $k \in K, 0 \leq t < 1$, $f(k,t) \in K$.

2) For all $x \in X$, and all sequences $\{t_i\} \to 1$, there exists a subsequence $\{t_{i_j}\} \to 1$ and $k \in K$ such that $\{f(x,t_{i_j})\} \to k$.

Does it follows that $K$ is homotopy equivalent to, or a deformation retract of, $X$?

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    $\begingroup$ A constant map satisfies your conditions. $\endgroup$ – Marc Hoyois Nov 14 '16 at 23:03
  • $\begingroup$ Do you want $f(x,0)=x$? $\endgroup$ – Denis Nardin Nov 14 '16 at 23:29
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No. Let $X$ be the topologist's sine curve, i.e. $X \subset \mathbb{R}^2$ is $\{0\} \times [-1,1] \cup \{(x,\mathrm{sin}(1/x)\mid x \in (0,1)\}$, and set $K = \{0\} \times [-1,1]$. Let $f_t$ fix $K$ for all times, while $f_t(x,\mathrm{sin}(1/x) = (x_t,\mathrm{sin}(1/x_t))$ where $x_t = \min(x,1-t)$. Then the homotopy $(f_t)_{t \in [0,1)}$ satisfies the hypotheses but $X$ is not homotopy equivalent to $K$ -- for instance $K$ is path-connected but $X$ is not.

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Can't you just take $X=[0,1]$ and $K=\{0,1\}$, with $$f(x,t)=\begin{cases}\frac{x}{1-t}&\mbox{if $t\leq 1-x$}\\ 1&\mbox{if $t\geq 1-x$},\end{cases}$$ so that $f(x,t)\to 1$ as $t\to 1$ unless $x=0$, in which case $f(x,t)\to 0$?

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  • $\begingroup$ much better example! $\endgroup$ – Tim Campion Nov 15 '16 at 21:14

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