It is well know that if $q=0$, then $H_q(\mathbb R)$ must be the integers (since $\mathbb R$ is path connected). In general, why is $H_q(\mathbb R)$ the trivial group for $q\geq 2$?
Why the $q^{\text{th}}$ homology group of $\mathbb R$ is the trivial group?
0
$\begingroup$
algebraic-topology
homology-cohomology
-
0$\Bbb R$ is contractible, so $H_q(\Bbb R) = H_q(\{0\}) = 0$ for $q \geq 1$. – 2017-01-24
1 Answers
2
$\mathbb{R}$ is contractible (i.e. homotopy equivalent to a point) and homology is a homotopy invariant, so
$$H_q(\mathbb{R}; \mathbb{Z}) \cong H_q(\{0\}; \mathbb{Z}) \cong \begin{cases} \mathbb{Z} &\ \text{if}\ q = 0\\ 0 &\ \text{otherwise}. \end{cases}$$
Note, $H_q(\mathbb{R}; \mathbb{Z}) = 0$ for $q = 1$ too.