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Monday, July 9, 2018

Homology Groups of $((S^1\vee S^1)\vee \bR P^2)\vee \bR P^3$

The following is a problem from an exam. We'll solve it supposing we can calculate the homology groups of $\bR P^n$ at least for $n\leq 3$ (Using cellular homology they aren't hard to calculate, and the homology of $\bR P^2$ can be easily calculated using simplicial homology). So, the reduced homology groups are just $H_n(\bR P^2)=\bZ_2$ if $n=1$ and $0$ otherwise. And $H_n(\bR P^3)=\bZ_2$ if $n=1$, $\bZ$ if $n=3$ and $0$ otherwise.

 Exercise. Find the fundamental group, the homology groups, and the cohomology ring mod 2 of the space $X$ given by pasting a circle, a real projective plane, and a three-dimensional real projective space to a circle along segments as follows (Ignore the blue $1-$cells):
Solution. The fundamental group is easy. $X$ is just the wedge sum $((S^1\vee S^1)\vee \bR P^2)\vee \bR P^3$ so, inductively, $\pi_1(X)\cong \bZ*\bZ*\bZ_2*\bZ_2$. Also, since the first homology group is just the abelianization of the fundamental group, we get $H_1(X)=\bZ\oplus\bZ\oplus\bZ_2\oplus\bZ_2$.

The homology group will also be computed inductively (We could use that the homology of a wedge sum is the direct sum of the homologies but we won't). To do this, take first just the wedge $S^1\vee S^1$. By the long homology sequence we have, for $n\geq 2$, an exact sequence of reduced homology groups:
$$H_n(S^1)\to H_n(S^1\vee S^1)\to H_n(S^1)$$
which is 
$$0\to  H_n(S^1\vee S^1) \to 0 $$
so $H_n(S^1\vee S^1)=\bZ\oplus \bZ$ if $n=1$ and $0$ otherwise.
In the same way, for $n\geq 2$ we get exact sequences:
$$H_n(\bR P^2)\to H_n ((S^1\vee S^1)\vee\bR P^2)\to H_n(S^1\vee S^1)$$
which are
$$0\to H_n ((S^1\vee S^1)\vee\bR P^2)\to 0$$
so for $n\geq 2$, $H_n ((S^1\vee S^1)\vee\bR P^2)=0$, and for $n=1$, $H_n((S^1\vee S^1)\vee\bR P^2)=\bZ\oplus \bZ\oplus \bZ_2$.

In a similar way, but with a slightly different result, for $n=2$ we get the exact sequence
$$H_2(\bR P^3)\to H_2(X)\to H_2(X/\bR P^3)$$
which makes $H_2(X)=0$. We also get, for $n=3$,
$$0\to H_3(\bR P^3)\to H_3(X)\to H_3(X/\bR P^3)$$
which is
$$0\to \bZ\to H_3(X)\to 0$$
which means that $H_3(X)\cong\bZ$.
For $n\geq 3$ we again get sequences of the form $0\to H_n(X)\to 0$.

Therefore, $H_n(X)=\bZ\oplus\bZ\oplus\bZ_2\oplus\bZ_2$ for $n=1$, $\bZ$ for $n=0,3$ and $0$ otherwise.

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