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, $X$?



Technical but elementary homotopy question
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.


