Fundamental group of $S^2$ with north and south pole identified Consider the quotient space obtained by identifying the north and south pole of $S^2$. I think the fundamental group should be infinite cyclic, but I do not know how to prove this. 
If it is infinite cyclic, would this and $S^1$ be an example of two spaces which have isomorphic fundamental groups but are not of the same homotopy type?
 A: Additionally, Hatchers book provides another way of answering this problem. We can realize $X$ (the space in question) as a CW complex, with $0$-skeleton a point, $1$-skeleton $S^1 \vee S^1$, say with fundamental group free on $a$ and $b$, and attach $2$ $2$-cells along $ab$, (or any other pair, depending on direction of paths and such). This leads to an isomorphism of the fundamental group of the $0$-skeleton $\mod (ab)$ with the fundamental group of $X$. Despite choice of how to glue the $2$-cells, we get $\langle a, b \mid ab \rangle$ which is infinite cyclic.
A: A sphere with two points identified is homotopy equivalent to the wedge of a sphere with a circle (this is proved in Hatcher's book, on page 11 of chapter 0). Thus, Van Kampen's theorem implies that the fundamental group is infinite cyclic. However, the second homology group is also infinite cyclic, so it's not homotopy equivalent to the circle.
A: Another way to see this is to use the theory of covering spaces. Consider the subspace $Y$ of $\mathbf R^3$ obtained by placing a copy of $S^2$ centered at $(n, 0, 0)$ for each even integer $n$ (I'll try to remember to add a picture later). The group $\mathbf Z$ acts on this space by translation, and the quotient $\mathbf Z\backslash Y$ is the space in question.

Since $Y$ is simply connected (can you prove this?), it follows from Proposition 1.40c of Hatcher that $\pi_1(\mathbf Z\backslash Y) \cong \mathbf Z$. I don't see a nice way of proving that the spaces aren't homotopy equivalent without using homology, as Chris does.
