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Let X be the quotient space of $S^2$ under the identifications $x\sim-x$ for $x$ in the equator $S^1$. Compute the homology groups $H_i(X)$. Do the same for $S^3$ with antipodal points of the equator $S^2 \subset S^3$ identified.

This is probably related to cellular homology. Thanks.

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    As Akhil said, I used one 0-cell, one 1-cell attached to the 0-cell in the obvious way and two 2-cells both attached to the previous with the antipodal map. So we get a sequence $0 \to^{d_3} Z^2 \to^{d_2} Z \to^{d_1} Z \to 0$ with $d_1 = 0$ and $d_2$ taking each generator of $Z^2$ to the generator of $Z$. That leaves us with $Ker d_1 = Z$, $Im d_2 = Z$, meaning $H_1=0$. Also, $Ker d_2 = 0$ and so $H_2 = 0$2011-01-10

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Let $X=S^2/\mathord\sim$ and,letting $S^2_+\subseteq S^2$ be the upper closed hemisphere in the sphere, let $X_+=S^2_+/\mathord\sim$ be the quotient of $S^2_+$ by the restricted equivalence relation. Now consider the long exact sequence in reduced homology for the pair $(X,X_+)$.

Using excision &c, show that the relative homology of $(X,X_+)$ is the same as that of the result of collapsing $X_+$ to a point, so that you get a $2$-sphere. On the other hand, $X_+$ is a projective plane, so you also know its homology. Now use the long exact sequence.

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    Could someone explain what the connecting map $\mathbb Z\rightarrow \mathbb Z/2\mathbb Z$ is? I am having trouble finding out what the non-trivial element in $H_1(\mathbb{RP^2})$ is. Thanks in advance.2014-06-23