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As I seem to have confirmed in this previous question, any compact 1-manifold is homeomorphic to the (possibly empty) disjoint union of circles. So my question reduces to essentially:

Does there exist any disjoint union of circles for which the only possible bordisms are not simply connected?

Intuitively, I think the answer is no, since it seems like we should always have a bordism that is essentially a collection of "tubes" extending from a "central hub", and that by contracting the tubes into the "central hub" we will always get something homeomorphic to a sphere, which is simply-connected. (See for example the picture below.)

enter image description here

However, I don't know how to prove or disprove this rigorously.

Also, please note that I do mean "the only possible bordisms are not simply-connected" -- it is easy to extend the above method to create non-simply-connected bordisms -- just put a "donut hole" in the center of the "central hub" to get a bordism homeomorphic to a torus (see the picture below). My question however is about whether it is ever the case that such a hole is necessary.

enter image description here

Note: The source for the first image is here, and the source for the second image is here.

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    $\begingroup$ The only simply connected compact surfaces are the sphere and unit disc. $\endgroup$ – user98602 Nov 19 '16 at 8:46
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    $\begingroup$ I may be understanding your question, but the "pair of pants" is not simply connected. $\endgroup$ – Danu Nov 19 '16 at 9:01
  • $\begingroup$ @Danu so are the two circles homologically equivalent to the one circle but not homotopically equivalent? The idea behind my question was to find a visually simple counterexample like this. Anyway intuitively it seems like a homotopy equivalence can be represented by a "cylinder/prism", whereas homology corresponds also to more complicated bordism relationships, assuming I'm not fundamentally misunderstanding something. $\endgroup$ – Chill2Macht Nov 19 '16 at 9:56
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    $\begingroup$ A (finite) disjoint union of copies of one connected manifold never has the same homology as one copy. Think about $H^0$! :) $\endgroup$ – Danu Nov 19 '16 at 10:11
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    $\begingroup$ You have $H^1(S^1 \sqcup S^1) = \mathbb{Z} \oplus \mathbb{Z} \neq H^1(S^1) = \mathbb{Z}$... $\endgroup$ – Najib Idrissi Nov 19 '16 at 12:02
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Yes. As Mike Miller says in a comment, the only simply connected compact surfaces are the 2-sphere $S^2$ and the disk $D^2$. The sphere can only be a bordism from $\varnothing$ to $\varnothing$, and $D^2$ is either a bordism from $\varnothing$ to $S^1$ or a bordism from $S^1$ to $\varnothing$, depending on the orientation.

So for any other combination of $1$-manifolds, none of the possible bordisms will be simply connected. For example a bordism $S^1 \to S^1$ is never simply connected.


The fact that the only simply connected compact surfaces are $S^2$ and $D^2$ follows from the classification theorem of compact surfaces.

  • If $\Sigma$ is a simply connected compact surface with no boundary, then it is orientable and thus it's the connected sum of $g$ toruses for some $g \ge 0$. But if $g > 0$ then we know the fundamental group, and it's not zero, therefore $\Sigma = S^2$.

  • If $\Sigma$ is a simply connected compact surface with boundary, then it is obtained by removing some number of disks from a closed surface, say $$\Sigma = \Sigma' \setminus \bigsqcup_{i=1}^k D^2.$$ But applying van Kampen's theorem several times, it follows that $\pi_1(\Sigma')$ is obtained as a quotient of $\pi_1(\Sigma) = 0$, hence $\Sigma'$ itself is simply connected. Therefore $\Sigma'$ is a sphere as we just saw.

    Removing one disk from $S^2$ yields $D^2$, which is simply connected. For $k > 1$, removing $k$ disks from $S^2$ yields a space homotopy equivalent to the wedge sum of $k-1$ circles, which isn't simply connected. Therefore $\Sigma' = D^2$.


Rereading your question, maybe you didn't really ask the question you wanted to ask. Indeed, the example you give isn't simply connected. Maybe you meant that when you fill out the disks in the bordism, then the result is simply connected?

In this case, then yes, given two $1$-manifolds $M$ and $N$, you can always find a bordism $\Sigma : M \to N$ such that when you fill out the circles in $\partial \Sigma$, then the result is simply connected. Just take a sphere $S^2$, and remove as many disks as there are circles in $M$ and $N$. The result is a bordism from $M$ to $N$, and if you fill out the holes then you get $S^2$ which is simply connected.

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  • $\begingroup$ Yes I think you are right, I was confusing circles with discs. Maybe more what I was trying to ask: does $H^2(D^2 \sqcup D^2) = H^2(D^2)$? And then does the resulting bordism ("filled in fork-shaped tube") have trivial fundamental group (obviously yes I think) but non-trivial 2nd homotopy group, or at the very least it isn't a homotopy between $D^2 \sqcup D^2$ and $D^2$ (since path connectedness is supposed to be a homotopy invariant I think, but $D^2 \sqcup D^2$ is not even connected, whereas $D^2$ is obviously a lot of nice things)? $\endgroup$ – Chill2Macht Nov 19 '16 at 13:16
  • $\begingroup$ See also my followup question on MathOverflow (although in hindsight I should have posted it on Math.SE): mathoverflow.net/questions/255096/… $\endgroup$ – Chill2Macht Nov 19 '16 at 18:52

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