Question about relative singular homology groups

Question

I know that the sphere $S^{\infty}$ is contractible, but why if $H$ is a Hilbert space then we have $$H_q(H,S^{\infty})=0, q\in \mathbb{N}?$$

Please help me

Thank you

Answer 1

Hint: We have a long exact sequence given as

$$\begin{array}{r}\cdots \to H_n(S^{\infty}) \to H_n(H) \to H_n (H,S^{\infty}) \stackrel{\delta}{\to} H_{n-1}(S^{\infty}) \to \cdots\\ \cdots\to \tilde{H}_0(S^{\infty}) \to \tilde{H}_0(H) \to H_0 (H,S^{\infty}) \to 0\end{array}$$

What does this reduce to after using the fact that $S^{\infty}$ and all Hilbert spaces are contractible?