Let $(X, \mathcal{T}_X)$ be a topological space.
Let $l$ be a loop so that $p \in X$ is the base point and $f: X \to X$ a continuous function, where $f \simeq id_X$ ($f$ and the identity function is homotopic).
Furthermore, let $X_l = \{x | x = l(t) ( 0 \leq t \leq 1) \}, X_{f \circ l} = \{ x | x = f \circ l(t) (0 \leq t \leq 1) \}$, and their topology defined by reative topology.
Are $X_l$ and $X_{f \circ l}$ homotopy equivalent?
This is not true. Let $X=\Bbb R^2$ and $f:\Bbb R^2\to \Bbb R^2$ be defined as $f(x)=0$ for all $x\in\Bbb R^2$. Define $H:\Bbb R^2\times [0,1]\to\Bbb R^2$ as $H(x,t)=tx$ for all $(x,t)\in\Bbb R^2\times [0,1]$. Then, $H:f\simeq \text{id}_{\Bbb R^2}$.
Next, let $l:[0,1]\ni t\longmapsto e^{2\pi it}\in \Bbb S^1$ be the loop based at $1\in\Bbb S^1$. Note that $X_l=\Bbb S^1$ and $X_{f\circ l}=\{0\}$. But, $\Bbb S^1$ is not homotopically equivalent to $\{0\}$ as $\pi_1(\Bbb S^1)=\Bbb Z$ but $\pi_1(\{0\})=\text{trivial group}$.