Inclusion is not homotopy equivalence

Question

How to prove that inclusion $i:(B^n,S^{n-1})\rightarrow (B^n,B^n\setminus\{0\})$ is not homotopy equivalence? ($B^n$ is closed unit $n$-ball and $S^{n-1}$ is boundary of $B^n$.)

Answer 1

Assume another map is $g$ s.t. $ig$ and $gi$ are homotopic to identity of pair.

Note that $g^{-1}(S^{n-1})$ is a closed set since $g$ is continuous. But $B^n$-${0}$ is open and is contained in $g^{-1}(S^{n-1})$. So $g^{-1}(S^{n-1})$ is $B^n$. Then $g$ restricted on $B^n$-${0}$ factors: $B^n$-{0}$\to B^n\to S^{n-1}$. Since $B^n$ is simply-connected, then $g_*$ induces trivial map from $H_{n-1}(B^n-{0})$ to $H_{n-1}(S^{n-1})$. So $g$ can not be a homotopy equivalence of pair. So is $i$.