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?
Best Answer
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 $H^1$.