If $\gamma:S^1\to X$ is a simple contractible loop, when can we say there must by a contraction, $H(s,t)$ such that $\gamma_t:s\mapsto H(s,t)$ is a simple loop for all $t<1?$
(1) It seems like you should be able to do this if $X$ is a manifold. (Of dimension $>1,$ but there are no retractable simple loops in manifolds of dimension $1.$) Maybe it requires dimension $>2?$
(2) I’d like to find a Hausdorff counterexample, or a proof that we can always find $H$ in those cases.
When there is a counterexample, the base case would be a quotient of the closed disk, with no more than one point on the boundary in each equivalence class. With $\sim$ defined cleverly enough, then $D^2/\sim$ with the loop $S^1\to D^2\to D^2/\sim $ would be an example.
A simple non-Hausdorff counterexample is $u\sim v$ if both $|u|,|v|<1/2.$ This quotient is non-Hausdorff, since an equivalence class is not closed. The singleton in $D^2/\sim$ consisting of the equivalence class of $0$ is open in $D^2/\sim.$ So you can’t cross $[0]$ to retract to another point - $\gamma_{t}^{-1}([0])$ must be an open set. And you can’t retract to $[0]$ point since $H^{-1}(\{[0]\})$ would have to be open, so must contain an $(s,t)$ with $t<1.$
If $X$ and $\gamma$ are a counterexample, then this means in particular that there is a $P:D^2\to X$with $P_{|S^1}=\gamma.$
If $X$ is Hausdorff, then when we define $u \sim_P v$ iff $P(u)=P(v),$ then $Y=D^1/\sim_P$ is homeomorphic to $P(D^2),$ since $Y$ is compact and $P(D^2)$ is Hausdorff.
If we can’t find a proper retraction in all of $X,$ we can’t find one in $P(D^2)$ and hence not in the case of $\gamma’:S^1\to D^2/\sim_P.$
So any Hausdorff counterexample gives us a Hausdorff counterexample of the form $D^2/\sim$ for some equivalence relation $\sim.$
Here's one manifold counterexample in three dimensions:
Let $X=\left(\mathbb{R}^2\setminus\{0\}^2\right)\times\mathbb{R}$ be $\mathbb{R}^3$ sans the $z$-axis, and consider the embedding $S^1\to X$ given in the following "knot diagram":
Where the $z$-dimension is suppressed. The loop is clearly contractible, but it cannot be contracted in the desired manner. To see this, suppose such an $H$ exists, and note that $\gamma_t$ is an unknot in $\mathbb{R}^3$ for all $t<1$, and there is an $\epsilon>0$ such that $\gamma_{1-\epsilon}$ is contained in an open ball disjoint from the $z$-axis. Adding a point at infinity, $H|_{S^1\times[0,1-\epsilon]}$ then becomes an isotopy which unlinks the Whitehead link, which is a contradition.