How does $\mathbb R^n\setminus \{0\}$ being simply connected follow as a corollary from $\mathbb S^{n-1}$ being a strong deformation retract?

Question

From Introduction to Topological Manifolds by Lee:

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We know $\mathbb S^{n-1}$ is simply connected for $n\ge 3$.

We know that since $\mathbb S^{n-1}$ is a strong deformation retract of $\mathbb R^n\setminus \{0\}$, then $r \circ \iota = \mathrm{Id}_{\mathbb S^{n-1}}$ and $\iota \circ r \simeq \mathrm{Id}_{\mathbb R^n\setminus \{0\}}$, where $\iota:\mathbb S^{n-1} \hookrightarrow \mathbb R^n\setminus \{0\}$ is inclusion.

We do not know homotopy invariance of $\pi_1$.

So, how does the corollary follow from the proposition?

Answer 1

Wow. This is just an oversight. Corollary 7.38 should have been delayed until after the statement of Theorem 7.40, which says that $\pi_1$ is homotopy invariant. I've added a correction.

Thanks for pointing this out.