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I came across this question while studying for a qualifying exam:

Prove that a closed orientable surface of genus $g \ge 1$ is not homotopy equivalent to the wedge $X \vee Y$ of two finite cell complexes both of which have nontrivial $H_1(\cdot;\mathbb Z)$.

I think that homology is not sensitive enough to solve this problem: You could take a surface of genus $g-1$ and wedge it with two circles and get the same homology as the surface of genus $g$.

But maybe the fundamental group is good enough: the standard presentation of the group for the surface of genus $g$ has one relation involving all the generators-- maybe we can observe that such a group can never be the free product of two non-trivial groups? But I'm not sure how to verify this. Or is there an easier way?

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See… for the statement about non-factorization of the surface group. – Mariano Suárez-Alvarez Aug 9 '12 at 22:08
Have you looked at the cohomology ring? – Mariano Suárez-Alvarez Aug 9 '12 at 22:10
Here's another way to do it: since the torus case is obvious, and we have the Kurosh subgroup theorem, it suffices to prove the fundamental group of a genus 2 surface (call it $G$) is not a free product. Using covering spaces, the only subgroups of $G$ are surface subgroups and free groups. The surface subgroups are higher genus (so require more generators), so the only way it could be a free product is as a free product of free groups, which is just a free group. But $G$ is not free. – user641 Aug 9 '12 at 22:13
Thank you for the ideas about free groups. I think the cohomolgy ring looks like a more elementary way to go. – paragon Aug 9 '12 at 22:45
@Steve D: Really dumb question: Is it obvious the fundamental group of a surface isn't free? (Of course, for $T^2$ it is obvious but what about the others? – Jason DeVito Aug 9 '12 at 22:56
up vote 1 down vote accepted

Thank you to Mariano Suárez-Alvarez. I think the cohomology ring will work, somthing like this:

We know that the cohomology ring of a genus $g$ surface has the property that any non-zero class in degree one has a class it can cup with to get a non-trivial element of $H^2$ (this could be from direct computation or by Poincare duality.) In the wedge sum, we notice that WLOG $X$ must have trivial $H^2$. Thus any two degree 1 class supported on $X$ cup to 0, and any class supported on $X$ cups to zero with a class supported on $Y$. Thus $X$ has trival $H^1$.

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This looks correct to me, but now your "1" is in the wrong place: superscripted instead of subscripted! I'd mention the universal coefficient theorem or somesuch to finish off the argument. – MartianInvader Aug 9 '12 at 23:59
Yes, since all the homology groups are free and finitely generated, the Universal coefficient theorem should say that the cohomology groups are isomorphic to the homology groups. – paragon Aug 10 '12 at 3:48

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