Is there a simply connected Hausdorff space $X$ which admits an embedding of $S^1$ that does not extend to a locally injective map $D^2 \to X$?

Question

It is a well-known (but not easy) fact that a (topological) embedding of the $0$-dimensional sphere $S^0 = \{-1, 1\}$ in a path connected Hausdorff space $X$ extends to an embedding of the $1$-dimensional closed unit disk $D^1 = [-1, 1]$; see Corollary 31.6 of [Willard] or the post Equivalence of Path-Connectedness and Arc-Connectedness for Hausdorff Spaces.

Let's assume $X$ is Hausdorff and simply connected, which means $X$ is path connected and every continuous map $S^1 \to X$ extends to a continuous map $D^2 \to X$. The existence of non-trivial knots in $S^3$ shows an embedding $S^1 \to S^3$ need not extend to an embedding $D^2 \to S^3$; this innocent statement encapsulates more difficulties than one might imagine. It seems too good to be true, so I guess it's also false in general that an embedding $S^1 \to X$ extends to a locally injective continuous map $D^2 \to X$. Question: What is an example of such $X$ and $S^1 \to X$?

For context, if there were no counterexamples, I think this would salvage a more restricted version of my previous question Can the real plane $\mathbb{R}^2$ with a topology strictly finer than the Euclidean topology be simply connected? (by providing a negative answer in the case that the topology is Euclidean outside of a bounded set).

Willard, Stephen, General topology, Mineola, NY: Dover Publications (ISBN 0-486-43479-6/pbk ). xii, 369 p. (2004). ZBL1052.54001.

Answer 1

Let me convert my comment to an answer. Let $X$ denote the presentation complex of the presentation $$ \langle a| a^2, a^3\rangle, $$ i.e. you attach two 2-cells to the circle $C=X^1$ by maps of degrees 2 and 3 respectively. Then $X$ is simply connected and contains a topological circle, namely $C$. I claim that there is no locally injective map $f: D^2\to X$ which restricts to a homeomorphism $\partial D^2\to C$. Suppose that such a map exists. Then one verifies that $A=f^{-1}(C)$ is a 1-dimensional submanifold of $D^2$, where $\partial D^2$ is one of the components. (This requires local injectivity of $f$ and a proof is a bit long.) Clearly, we cannot have $A= \partial D^2$ (for otherwise, $f(D^2\setminus A)$ would be one of the open 2-cells of $X$, which is impossible). Then $D^2\setminus A$ contains a non-simply connected component $B$. Then $f(B)=e$ is one of the two open 2-cells in $X$. Furthermore, the restriction $f: B\to e$ is a proper local homeomorphism. Hence, it is a covering map. But $e$ is simply-connected, hence, this restriction is a homeomorphism, contradicting the fact that $B$ is not simply-connected.