Simply connected closed subspace of Banach space is a retract

Question

Suppose that $K$ is a simply connected and closed subset of an infinite-dimensional Banach space $B$. Then is $K$ necessarily a retract of $B$?

I can't seem to find a counter example..

Answer 1

This community wiki solution is intended to clear the question from the unanswered queue.

Moishe Kohan has answered the question in his comments.

It is obvious that a simply connected and closed subset $K$ of a finite-dimensional Banach space $F$ is not necessarily a retract of $F$. As an example take the unit sphere $S = \{ x \in F \mid \lVert x \rVert = 1 \}$ in $F$. If $\dim F =n > 1$, then $S$ is simply connected (it is homeomorphic to the standard unit sphere $S^{n-1}$ in $\mathbb R^n$).

Now let $B$ be an infinite-dimensional Banach space. Consider a finite-dimensional linear subspace $F \subset B$ with $\dim F > 1$. $F$ is closed in $B$. Then take $K \subset F$ which is not a retract of $F$. Then trivially $K$ cannot be a retract of $B$.

Although the unit spheres $S$ are no retracts of $F$ in the finite-dimensional case, they are at least neigborhood retracts which means that $S$ is a retract of an open neigborhood $U$ of $S$. Here is an example for which not even this is true: The Warsaw circle https://de.wikipedia.org/wiki/Datei:Warsaw_Circle.png , How to show Warsaw circle is non-contractible? is a compact simply connected subset of the plane $\mathbb R^2$, but it is not a retract of any of its open neigborhoods.

Answer 2

1. Each Banach space is contractible.

2. Each retract of a contractible space is contractible.

3. Each Banach space of dimension $> 1$ contains a compact simply connected subset which is not contractible.