Let $\mathbb{H}^n = \{x = (x_1,\dots,x_n) \in \mathbb{R}^n \mid x_n \geq 0 \}$. i.e. the half plane. Let $y \in \mathbb{R}^{n-1} \times \{0\}$ be a point on the boundary.
How do I show that $H_q(\mathbb{H}^n \mid y) = 0$ for all $q$?
Attempt: we have a long exact sequence,
$$\cdots \to H_q(\mathbb{H}^n - y) \to H_q(\mathbb{H}^n) \to H_q(\mathbb{H}^n \mid y) \to H_{q-1}(\mathbb{H}^n -y) \to H_{q-1}(\mathbb{H}^n) \to \cdots.$$
Now, $H_q(\mathbb{H}^n) = 0$ for all $q$ since the half plane, like all of $\mathbb{R^n}$, is contractible. So, $H_q(\mathbb{H}^n \mid y) \cong H_{q-1}(\mathbb{H}^n - y)$.
However, $H_{q-1}(\mathbb{H}^n - y) = 0$ for all $q$ since $\mathbb{H}^n - y$ is also contractible?