homology groups of the quotient of $\mathbb{S}^2$ obtained by identifying north and south poles to a point

Question

Compute the homology groups of the quotient of $\mathbb{S}^2$ obtained by identifying north and south poles to a point.

I have already calculated the homology groups in one way, but I would like to calculate the homology groups also using the mayer vietoris sequence, so I am trying to find two open $U\subset X=\mathbb{S}^2/{\sim}$ and $V\subset X$ such that $X=U\cup V$, I have initially taken $U=X-\{(0,0,1)\}$ and $V=X-\{(1,0,0)\}$, but I do not know how these spaces look, could someone tell me which spaces are homotopic $U$ and $V$? Thank you.

Answer 1

You can draw your space in the following way:

It looks a bit like a nice croissant. The $U$ and $V$ you have given are (as Eric Wofsey points out in the comments) your space $X$ without the identification point and $X$ with any other point removed, respectively. While $U$ is an open annulus, I don’t see a space that $V$ is homotopic to that has homology groups I know immediately. Thus, I would suggest a different choice.

Here, the blue portion in the picture above should be your $U$ and the red portion should be your $V$. Thus $U$ is an open annulus which is homotopic to a circle, and $V$ is the wedge sum of two disks which is homotopic to a point.

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This will help you formally compute the homology groups. That said, I think it can often be helpful to “find” the answer before you “know” the answer. We can do this by drawing your space is as the following quotient of a torus:

where we identify all points on the pink circle. Since the pink circle is a generator of one $\mathbb{Z}$ in the first homology of the torus, this is squashed to $0$ in the first homology of our space — and nothing else has changed. Thus, we should expect that since

$$H_n(T^2) = \begin{cases} \mathbb{Z} & n=0,2 \\ \mathbb{Z}^2 & n=1 \\ 0 & n\ge 3\end{cases}$$

we ought to have

$$H_n(X) = \begin{cases} \mathbb{Z} & n=0,1,2 \\ 0 & n\ge 3\end{cases}$$

Answer 2

Regard $X$ instead as $S^2 \coprod I/(0,0,\pm 1) \sim (\pm 1)$, where $I=[-1,1]$. This is $S^2$ with a line segment joining the north and south pole. Up to homotopy equivalence, this is your space. Using your neighborhoods, you will find that $U \cap V \cong S^1\times I \coprod I$ (of course up to homotopy we can push down to get $S^1 \coprod pt$. Then by Mayer Vietors and the LES of reduced homology, we get

$$0 \to \tilde{H}_2(X) \to \tilde{H}_1(U\cap V ) \to 0 \to \tilde{H}_1(X) \to \tilde{H}_0(U \cap V) \to 0$$

From this, you will deduce that $H_2(X)=\mathbb Z$ and $H_1(X)=\mathbb Z$ and zero everywhere else.

Actually if you stop and ponder what $X$ is, you might notice that $I$ might be outside of the sphere, and you can in fact contract the endpoints of $I$ to a point, and note that $X=S^2 \vee S^1$ up to homotopy equivalence.

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If you are insistent on not changing the problem at all, then you can prove that $U \cap V$ is still $(S^1 \times I) \coprod pt$ set theoretically, where $pt=\{(0,0,1),(0,0,-1)\}$ Since $pt \in U,V$ and the rest is the usual intersection.