Is $(I^n, \partial I^n)$ homotopy equivalent to $(R^n, R^n\setminus \left\{0\right\})$

Question

I was asked for a homework to show that the pair $(I^n, \partial I^n)$ is homotopy equivalent to $(R^n, R^n\setminus \left\{0\right\})$, where a relative homotopy between two maps $f, g: (X, A)\rightarrow(Y, B)$ is a continuos map $F:(X\times I, A\times I)\rightarrow(Y, B)$ (which means that $F(a, t)\in B$ $\forall a\in A$, $\forall t\in I$) such that, as usual, $F(-, 0)= f$ and $F(-, 1) = g$.

For $n=2$, my idea was to replace $(I^2, \partial I^2)$ by $(D^2, S^1)$, since they are homeomorphic, and to use as candidates for quasi inverses the constant maps $f:(D^2, S^1)\rightarrow(R^2, R^2\setminus\left\{0\right\})$, $g: (R^2, R^2\setminus\left\{0\right\})\rightarrow(D^2, S^1)$ both with value $(1,0)$. I think that the map $F$

$$(r\cos\theta, r\sin\theta, t)\mapsto \begin{cases} r(\cos(-2t\theta)\cos\theta - \sin\theta \sin(-2t\theta), \cos\theta \sin(-2t\theta) + \sin\theta \cos(-2t\theta) ) & t\in[0,1/2] \\ (r(2-2t) + 2t - 1, 0) & t\in[1/2,1] \end{cases}$$

should work as a relative homotopy between $f\circ g$ and $id_{R^2}$ and, up to a restriction, as a relative homotopy between $g\circ f$ and $id_{D^2}$ because, if I start at $t=0$ from a point in $S^1$, the homotopy just makes it go around the circle until, at $t=1$, it arrives at the point $(1,0)$, so $F(x, t)\in S^1$ $\forall x\in S^1$, $\forall t\in I$.

My intuition is that, for $n>2$, up to making several rotations around the Cartesian axes, a similar homotopy should work, even though I'm not sure about how it should be written explicitly. So my questions are:

1. Is my proof for $n=2$ correct?
2. Can I do something similar for $n>2$? If so, how should the homotopy be defined?
3. How do I deal with the case $n=1$?

Answer 1

$\require{AMScd}$It seems to me such a homotopy equivalence of pairs can't exist: assume the contrary (in fact less: assume there is a map of pairs $(g_0, g_1)$ that makes the square $$ \begin{CD} R^n\setminus\{0\} @>g_0>> \partial D^n \\ @VVV @VVV \\ R^n @>>g_1> D^n \end{CD} $$ commute). Then $g_1^\leftarrow(\partial D^n)$ is closed in $R^n$; the commutativity entails that it contains $R^n\setminus\{0\}$, so it contains its closure, the whole $R^n$. But then $g_1$ factors through $\partial D^n$, which is impossible if you further assume that the pair is a homotopy equivalence (precompose with an obvious homotopy equivalence $R^n\simeq D^n$: this would violate Brouwer theorem).

Answer 2

Let $f : (X,A) \to (Y,B)$ be a homotopy equivalence of pairs and $f_{rel} : A \to B$ be the map obtained by restriction. Then $f_{rel}$ is obviously a homotopy equivalence of spaces. Assume there exists a homotopy equivalence of pairs $h : (\mathbb{R}^n,\mathbb{R}^n \backslash \lbrace 0 \rbrace) \to (D^n,S^{n-1})$. Then $h_{rel} : \mathbb{R}^n \backslash \lbrace 0 \rbrace \to S^{n-1}$ is a homotopy equivalence. We must have $h(0) \in S^{n-1}$, otherwise $h : \mathbb{R}^n \to D^n$ would not be continuous (since each neighborhood of $0$ contains points of $\mathbb{R}^n \backslash \lbrace 0 \rbrace$ which are mapped to $S^{n-1}$). Hence we can write $h_{rel} = h \circ i$ with inclusion $i : \mathbb{R}^n \backslash \lbrace 0 \rbrace \to \mathbb{R}^n$. This means that $h_{rel}$ is null-homotopic because $\mathbb{R}^n$ is contractible. We conclude that $S^{n-1}$ must be contractible (because a null-homotopic map is a homotopy equivalence if and only if the spaces involved are contractible). But $S^{n-1}$ is not contractible; this disproves the above assumption.