TITLE: Is the sphere with a diameter homotopy equivalent to a surface?
QUESTION [7 upvotes]: This is for a homework problem:

Take the unit sphere $\mathbb{S}^2$ and join the north and south poles with a line segment. Is the resulting space homotopy equivalent to a surface?

Intuitively, I think the answer is no, because I can't think of any way to deform the poles to become locally Euclidean. I've noted that the fundamental group is $\mathbb{Z}$ and that the space can be deformed into a sphere with the poles identified, or into a torus with a disc in the middle. But I don't know how to use any of this.

REPLY [5 votes]: Call this space $X$.  Since $H_2(X)$ is nontrivial, if $X$ is homotopy equivalent to a surface $\Sigma$, $\Sigma$ must be compact and orientable.  By the classification of compact surfaces, no such surface exists with fundamental group $\mathbb{Z}$.
Alternatively, if you don't want to use the classification of surfaces, you can just use Poincare duality: the cohomology of $X$ does not satisfy Poincare duality.  Indeed, $H^1(X)\cong\mathbb{Z}$, and by anticommutativity $x^2=0$ if $x\in H^1(X)$ is a generator, so the cup product pairing on $H^1(X)$ cannot be nondegenerate.  
For yet another argument, take the double cover $Y$ of $X$ corresponding to the subgroup $2\mathbb{Z}\subset\mathbb{Z}$.  Then $H_2(Y)=\mathbb{Z}^2$, which cannot be $H_2$ of a surface.  But $Y$ is homotopy equivalent to the corresponding double cover of $\Sigma$, which is a surface.