8

Let $X$ be a connected space homeomorphic to a finite simplicial complex. If there is an embedding $i: S^1 \hookrightarrow X$ which has a retract $r: X \rightarrow S^1$, then necessarily the first Betti number $b_1(X)$ is nonzero. Is the condition $b_1(X) \neq 0$ sufficient for a circle retract to exist?

Cihan
  • 2,426
  • 12
  • 30
  • It certainly depends on the embedding. See [here](https://math.stackexchange.com/questions/61366/surface-of-genus-g-does-not-retract-to-circle-hatcher-exercise) – H1ghfiv3 Mar 05 '18 at 20:09
  • @BerniWaterman I didn't mean to fix an embedding. The question is: If $b_1(X) \neq 0$, is there _any embedding_ whatsoever from the circle which has a retraction? – Cihan Mar 05 '18 at 22:22
  • I see... interestingly enough, this statement cannot be generalized to higher dimensional spheres, and it is easy to find counterexamples: For $n \geq 2$, take $X$ to be a closed, oriented hyperbolic $n$-manifold. Then $b_n(X) > 0$, but $X$ certainly does not admit an $S^n$ retract. – H1ghfiv3 Mar 06 '18 at 12:19

3 Answers3

4

As mentioned in Cihan's answer, this is a result of Borsuk. The proof is divided into two papers (see [1, Théorème 30] and [2, Korollar 11]). A proof of this result (for Peano continua and rather difficult to read) appears in Kuratowski's book [3, S 57, III, Theorem 4].

In Section 1.1.3 (pages 8-11) of my thesis [4], I provide a proof for arbitrary (not necessarily finite) simplicial complexes which is based on Kuratowski's proof.

I sketch here a different way to prove this result for a finite simplicial complex $X$. First we take a 1-cycle in $Z_1(X)$ which represents a generator of a direct summand of $H_1(X)$ isomorphic to $\mathbb{Z}$. Then we need to take a representative of this cycle which is a simple closed curve. If $X$ is not a surface and $X$ has no local separating points (a point $x\in X$ is a local separating point if there is a connected open neighborhood $U\ni x$ such that $U-x$ is disconnected) it is possible to do this. The main idea is to consider three 2-simplices with a common edge and to use this to remove intersections one by one. It is easy to reduce to the case in which $X$ has no local separating points. Finally, surfaces are managed using the classification.

In every proof I know, it is necessary to subdivide the complex at some point. I do not know if it is always possible to find a subcomplex of the original complex homeomorphic to $S^1$ which is a retract.

[1] K. Borsuk. Quelques théorèmes sur les ensembles unicohérents, Fund. Math. 17 (1931), no. 1, 171–209.

[2] K. Borsuk. Über die Abbildungen der metrischen kompakten Räume auf die Kreislinie, Fund. Math. 20 (1933), no. 1, 224–231.

[3] K. Kuratowski. Topology vol. ii, Academic Press, 1968.

[4] I. Sadofschi Costa. Fixed points of maps and actions on 2-complexes, PhD thesis, Universidad de Buenos Aires, 2019. Available at http://cms.dm.uba.ar/academico/carreras/doctorado/tesisSadofschi.pdf.

2

Interesting question. I believe it is sufficient, let me sketch a proof.

Since $$b_1(X) = \text{rank}(H_1(X;\mathbb{Z})) = \text{rank}(H^1(X;\mathbb{Z})) $$ it follows from $b_1(X) \ne 0$ that $H^1(X;\mathbb{Z})$ is nontrivial. Consider the canonical bijection $[X,S^1] \approx H^1(X;\mathbb{Z})$, which associates to each $f : X \to S^1$ the pullback via $f$ of the fundamental cohomology class of $S^1$, denoted $f^*(d\theta) \in H^1(X;\mathbb{Z})$ (if I may abuse notation). Let $f : X \to S^1$ be any element whose pullback class $f^*(d\theta)$ is a basis element of $H^1(X;\mathbb{Z})$. It follows that there exists a continuous map $\sigma : S^1 \to X$ such that the $f^*(d\theta)$ evaluates to $+1$ on $\sigma$, and so $f \circ \sigma : S^1 \to S^1$ is homotopic to the identity.

Claim: One can homotope $f$ and $\sigma$ so that $f \circ \sigma$ is equal to to the identity on $S^1$.

Once this claim is proved, it follows that $\sigma \circ f : X \to \sigma(S^1)$ is a retract onto a circle.

Let me sketch a proof of the claim. First, by homotoping $\sigma$, we may assume that $\sigma$ is a concatenation of 1-cells of $X$, $\sigma = e_1 \cdots e_K$. Next, we may assume that for each 1-cell $e_k$, the restriction $f | e_k$ goes around $S^1$ monotonically, either forward, backward, or constant; this is true by the homotopy extension lemma. And then we can assume that $f$ is not constant on each $e_k$, or else we may do a small homotopy of $f$ near an endpoint of $e_k$, and then homotope $f$ to be non constant on $e_k$, and then apply homotopy extension.

Let $P$ be the number of local maxima of $f \circ \sigma$, which is equal to the number of local minima. The proof now proceeds by induction on $P$. If $P=1$ it should be clear that $f \circ \sigma$ is a homeomorphism, and then by composing with an isotopy of $\sigma$ we can make $f \circ \sigma$ the identity. If $P \ge 2$ then (after a cyclic permutation) we may assume that the restrictions of $\sigma$ to $e_K$ wraps negatively, to $e_1,...,e_J$ wrap positively, and to $e_{J+1}$ wraps negatively. Now we do a subsidiary induction on $J$: to reduce $J$, homotope $\sigma \mid e_{J-1} e_J$ to be monotonic, and apply homotopy extension.

Lee Mosher
  • 99,349
  • 6
  • 62
  • 136
  • In particular, this implies that for any (finite) CW-complex $X$ with $b_1(X) > 0$, there is always an element $[\gamma] \in \pi_1(X)$ of infinite order that can be represented by a simple closed curve... interesting... – H1ghfiv3 Mar 05 '18 at 22:31
  • Actually there are problems with this proof. The proof is invalid, for example, if $X$ is a wedge of two circles $a,b$ and $\sigma = ab$. I think it can be repaired (in the wedge of two circles, $\sigma$ should be $a$ or $b$, just not $ab$ or something more complicated). But I have to think about it more and I'm out of time for the day. – Lee Mosher Mar 06 '18 at 03:17
1

The answer is yes, via results of Borsuk and Čech from the 30’s. The generality where the sufficiency holds is remarkable, but the route is rather undirect this way. As I’m answering my own question, I will wait for some time before accepting this as an answer. There should be a shorter and self-contained proof, at least in the triangulable case, along the ideas Lee Mosher has outlined in the previous answer.

Definition: A space $X$ is called a Peano continuum if it is compact, connected, metrizable, and locally connected.

This terminology is due to the famous Hahn-Mazurkiewicz theorem, which says that every Peano continuum can be realized as a continuous image of the unit interval $[0,1]$.

Definition: A connected topological space $X$ is called unicoherent if for any decomposition $X = A \cup B$ with $A$ and $B$ closed and connected, the intersection $A \cap B$ is also connected.

Theorem: Let $X$ be a Peano continuum. The following are equivalent:

  1. The first Betti number $b_1(X)$ is nonzero.
  2. $X$ is not unicoherent.
  3. $X$ admits the circle $S^1$ as a retract.

Čech proved 1 $\Rightarrow 2$ as Theorem B here. Borsuk proved 2 $\Rightarrow$ 3 as Theorem 30 here. And 3 $\Rightarrow$ 1 is okay.

Cihan
  • 2,426
  • 12
  • 30
  • I believe that $b_1(X) \neq 0 \Leftrightarrow X$ has $S^1$ retract should also hold for not necessary compact, but reasonably well behaved, simplical complexes (for example, smooth manifolds). – H1ghfiv3 Mar 07 '18 at 11:21
  • Also, do you know if there exists english translations of those papers ? – H1ghfiv3 Mar 07 '18 at 19:07
  • @BerniWaterman I don't, unfortunately. But, to my surprise (I don't know German or French), it wasn't terribly difficult to find my way through those papers using Google translate. – Cihan Mar 07 '18 at 21:21