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There is a short argument using Zorn's lemma and the compactness of $[0,1]$, that shows every manifold must have maximal open simply connected subspaces.

However, I am wondering if it is necessarily the case that these subspaces are dense. It seems quite obvious to me that they must be dense, but am having some difficulty with the proof. Is this true, or do I lack the imagination to come up with a counterexample?

As Nate pointed out, I should have required the manifold to be connected.

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    $\begingroup$ At a minimum, you want to assume $M$ is connected, right? $\endgroup$ Sep 3, 2015 at 4:14
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    $\begingroup$ @JohnMa That was what I tried originally, however I run into issues, that I couldn't seem to fix. For instance, say that there is a 'slit' cut out of the open meaning some line segment removed from the boundary. How do we know that you don't say accidentally, close up part (but not all) of the slit and end up with something that is not simply connected. Now it seems these "slits" can't crop up in these maximal open sets, but I'm not sure how to show it. $\endgroup$
    – Bill Trok
    Sep 3, 2015 at 4:48
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    $\begingroup$ @JasonDeVito: Given an increasing sequence of subspaces $U_1 \subset U_2 \subset \dots$, call their union $U$. If every compact subset of $U$ is contained in some $U_i$ then $\lim \pi_1(U_i) = \pi_1(U)$. Indeed this holds for all homotopy groups or homology groups. Proof: there is a canonical map from the left side of that equality to the right side. Surjectivity: any loop in $U$ lies in some $U_i$ by assumption. Injectivity: if a loop in $U_i$ was null-homotopic in $U$, that null-homotopy would lie in some $U_j \supset U_i$, by the same compactness assumption, which is true for $U_i$ open. $\endgroup$
    – user98602
    Sep 3, 2015 at 18:57
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    $\begingroup$ @JasonDeVito Given a map $S^1\to M$ the image is compact so as it is covered by the chain it is covered by some element in the chain, and that element in the chain must be simply-connected by assumption. $\endgroup$ Sep 3, 2015 at 18:58
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    $\begingroup$ I'd just like to point out that the "manifold" condition would be completely essential if it's true, even very nice spaces (manifold with a single singularity) wouldn't work: look at $S^1 \vee [0,1]$, then $S^1$ minus the point at which $[0,1]$ is attached is a maximal simply connected open subset which isn't dense (I've been pulling my hair out thinking about this problem for hours and the only conclusion I've reached is that I've no idea if it's true). $\endgroup$ Sep 4, 2015 at 8:27

1 Answer 1

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The following answers a weaker form of your question, namely, is it true that every connected topological manifold contains an open dense simply-connected subset? The answer to this question is positive and even more is true:

Theorem. Every connected topological $n$-manifold contains an open dense subset homeomorphic to $R^n$.

This follows from a theorem by R.Berlanga "A mapping theorem for topological sigma-compact manifolds", Compositio Math, 1987, vol. 63, 209-216.

Berlanga's theorem generalized an earlier work by M.Brown, who proved the same theorem for compact topological manifolds. (The case of triangulated manifolds is easy but serves as a guideline for proofs in the topological category.)

Berlanga's theorem does not answer, however, the question if every maximal simply-connected open subset is dense in $M$.

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