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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..

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    I'm fairly certain that the unit sphere $K:={x\in B| ||x||=1}$ is not a retract of $B$, for reasons similar to the usual theorem that there is no retract from the closed unit ball to the unit sphere in finite dimension. $K$ is simply connected by the Van Kampen Theorem. – WoolierThanThou Jul 26 '19 at 18:30
  • Oh, right. I'm silly. –  Jul 26 '19 at 20:44
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    @WoolierThanThou But in contrast to the finite-dimensional case the unit sphere in $B$ is contractible. See e.g. https://www.ams.org/journals/proc/1983-088-03/S0002-9939-1983-0699410-7/S0002-9939-1983-0699410-7.pdf. – Paul Frost Jul 26 '19 at 21:49
  • Instead of the hypersphere, just take a 2-dimensional sphere inside your Banach space $B$. It will not be a retract of $B$. A more interesting question is if every contractible closed subspace is a retract. This is still false. But if you assume, in addition that the subset is finite-dimensional and locally contractible then indeed it will be a retract. https://en.wikipedia.org/wiki/Retract#Absolute_neighborhood_retract_(ANR) – Moishe Kohan Jul 27 '19 at 00:52
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    @WoolierThanThou Actually, the statement that you are fairly certain about is false: If $H$ is an infinite-dimensional Hilbert space, then it retracts to the hypersphere in $H$. The reason is that $H$ is homeomorphic to $H \setminus {0}$ by a homeomorphism fixing the unit hypersphere pointwise. https://math.stackexchange.com/questions/1356622/existence-of-a-continuous-mapping-that-maps-the-closed-unit-ball-onto-its-exteri/1620507#1620507 – Moishe Kohan Jul 27 '19 at 01:05
  • Well, now I'm silly. – WoolierThanThou Jul 27 '19 at 07:15
  • This is very interesting. I never knew this. Thank you :) –  Jul 28 '19 at 11:58

2 Answers2

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  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.

Paul Frost
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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.

Paul Frost
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