Algebraic Topology – Computing Homology Groups from Fundamental Group

algebraic-topologyhomology-cohomologysolution-verification

I would like some input on my calculations of the homology groups of a compact oriented 3-manifold, $X$, with $\pi_1(X) \cong \mathbb{Z} / 7$.

First note that X admits a cw structure of dimenson $3$, which implies that $H_i(X) \cong 0 \; \forall \; i \geq 4$. Moreover a connected manifold is also path connected, hence $H_0(X) \cong \mathbb{Z}$.

One can prove that for every orientable n-manifold $Y$ we have $H_n(Y) \cong \mathbb{Z}$. Hence $H_3(X) \cong \mathbb{Z}$.

Further one can obtain the first homology group as the abelianization of $\pi_1(X)$. However, $\mathbb{Z}/7$ is already abelian. Thus, $H_1(X) \cong \mathbb{Z}/7$.

To obtain the second homology group I figured I could apply Pioncaré Duality and observe that $H^1(X) \cong H_2(X)$. Hence I need the first cohomology group of X, which one can obtain through UCT for cohomology

\begin{align}
H^1(X;\mathbb{Z}) &\cong Hom(H_1(X; \mathbb{Z}), \mathbb{Z}) \oplus Ext^1_\mathbb{Z}(H_0(X), \mathbb{Z}) \\
H^1(X;\mathbb{Z}) &\cong Hom(\mathbb{Z}/7, \mathbb{Z}) \oplus Ext^1_\mathbb{Z}(\mathbb{Z}, \mathbb{Z}) \\
H^1(X) &\cong 0 \oplus 0 \cong 0 \cong H_2(X)
\end{align}

I obtain the following homology groups:
\begin{align}
H_0(X) &\cong \mathbb{Z} \\
H_1(X) &\cong \mathbb{Z}/7 \\
H_2(X) &\cong 0 \\
H_3(X) &\cong \mathbb{Z} \\
H_i(X) &\cong 0 \; \forall \; i \; \geq 4
\end{align}

Any comments are appreciated:)

Best Answer

First note that X admits a cw structure of dimenson $3$, which implies that $H_i(X) \cong 0 \; \forall \; i \geq 4$.

This is correct, although I'm of the opinion that it is much easier to show that the homology of a manifold vanishes above its top dimension than it is to show that $3$-manifolds admit CW structures.

Moreover a connected manifold is also path connected, hence $H_0(X) \cong \mathbb{Z}$.

Assuming you meant to assume your manifold is connected (it is not stated in your first paragraph), this is correct.

One can prove that for every orientable n-manifold $Y$ we have $H_n(Y) \cong \mathbb{Z}$. Hence $H_3(X) \cong \mathbb{Z}$.

The correct hypothesis is that $Y$ be orientable, compact and connected.

The rest (and hence your conclusion) is correct.

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