how to prove $L(m)$ is a orientable 3-manifold and compute $H^{*}(L(m))$

Question

Suppose $S^3=\{(z_1,z_2)\in \mathbb C^2\mid|z_1|^2+|z_2|^2=1\} $

there is a $\mathbb Z/m$ group action on $S^3$:
$$\phi:\mathbb Z/m \times S^3 \rightarrow S^3:\phi(k,(z_1,z_2))=e^{\frac{2k\pi }{m}\dot i} (z_1,z_2)$$ Let $L(m)$ be the quotiont space:$S^3/{\sim}$ :where $e^{\frac{2k\pi }{m}\dot i}(z_1,z_2)\sim(z_1,z_2)$

This is a standard computation; see Hatcher, for example. – anomaly Jun 16 '18 at 16:29 — anomaly, Jun 16 '18 at 16:29

score 1 · Answer 1 · answered Jun 16 '18 at 16:46

You can observe that the action of $\mathbb{Z} /m$ on $S^3$ is properly discontinuous and so the projection map is a covering map. In this way you have that $S^3/ \sim $ inherits the property to be locally Euclidean of dimension 3 (because $S^3$ is a real 3-manifolds). So you must verify that is also $T_2$ and second countable. For the second countable you can use that the image of a second countable space with respect to a countinuos surjective map is second countable. The property $T_2 $ is more difficult. You can prove that if G is a finite group that acts to a compact and $T_2$ topological space than the projection map is always closed and consequently the quotient space is $T_2$.
So the quotient space is a real 3-manifold.

For the calculus , for example, of the first Homological group of $Y=S^3/ \sim $ you can use the theorem that states that if $Y$ is a path-connected topological space such that his first fondamental group is abelian than

$H_1(Y)\cong \pi_1(Y)$

So you can use the property of covering map to obeserve that $\pi_(Y)\cong \mathbb{Z} /m$ because $S^3$ is also simply connected.

how to prove $L(m)$ is a orientable 3-manifold and compute $H^{*}(L(m))$

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