The $n$-dimensional cube modulus its boundary is homeomorphic to an $n$-dimensional sphere

Question

Let ${I}^{n}$ denote the $n$-dimensional cube, $\partial{I}^{n}$ be its boundary and ${S}^{n}$ denote the $n$-dimensional unit sphere.

Now for a pointed space $(X,x_{0})$ the $n^{th}$$homotopy$ $group$ of $X$ is :

1. The set of homotopy classes of maps $f:({I}^{n},\partial{I}^{n}) \rightarrow (X,x_{0})$ such that the boundary $\partial{I}^{n}$ goes to the basepoint $x_{0}$ where a homotopy $f_{t}$ satisfies $f_{t}(\partial{I}^{n})=x_{0} \ \forall \ t\in[0,1]$.

Alternatively;

2. The set of homotopy classes of maps $g:({S}^{n},s_{0}) \rightarrow (X,x_{0})$.

Now this holds as $I^{n}\backslash\partial{I}^{n}$ is homeomorphic to $S^{n}$.

This is where I struggle. Is it true to say that from the second definition we would send $s_{0}$ to $x_{0}$ ( hence one point to one point) but in the first we would be sending the set of boundary points to a unique point $x_{0}$. This many-to-one and one-to-one difference in the mapping defined in the first and the second definition respectively, makes it hard for me to see that the two are alternative to each other since the difference in mapping makes it seem like two contrasting processes, even if I understand that it is an obvious consequence from the homeomorphism between the $I^{n}\backslash\partial{I}^{n}$ and $S^{n}$. It would seem more natural for me if in definition 1 the mapping was from $f:({I}^{n}\backslash\partial{I}^{n},i_{0}) \rightarrow (X,x_{0})$ is that not possible?

Answer 1

You write that you are concerned that the function $I^n \mapsto S^n$ is many-to-one on the subset $\partial I^n$ and one-to-one on the rest of $I^n$. Well, that's exactly what the concept of quotient topology and quotient map are designed to do: it's not $I^n$ itself which is homeomorphic to $S^n$, its the quotient space $I^n/\partial I^n$ in which $\partial I^n$ is regarded as a single element, and each other element is regarded as a single point of $I^n - \partial I^n$.

To formalize this, the topological space denoted $I^n / \partial I^n$ refers to the quotient space of $I^n$ with respect to the decomposition of $I^n$ into the following collection of subsets: $$\mathcal D = \{\partial I^n\} \cup \{\{x\} \mid x \in I^n - \partial I^n\} $$ In words, each decomposition element of $\mathcal D$ is either the single subset $\partial I^n$ or a one-point subset $\{x\}$ as $x$ varies over $I^n - \partial I^n$. The underlying set of the quotient space $I^n / \partial I^n$ is simply the set $\mathcal D$ itself. The quotient map $q : I^n \to I^n / \partial I^n$ assigns to each $x \in I^n$ the unique element $q(x) \in \mathcal D$ such that $x \in q(x)$. The quotient topology is the strongest topology on $I^n / \partial I^n$ such that the function $q$ is continuous; equivalently, a subset $U \subset I^n / \partial I^n$ is open if, and only if, $q^{-1}(U) \subset I^n$ is open.

If we define the base point of $I^n / \partial I^n$ to be $i_0 = \{\partial I^n\}$, then to show that $(I^n / \partial I^n,i_0)$ is homeomorphic to $(S^n,s_0)$, one may use the universality property of quotient maps: it suffices to construct a surjective function $p : I^n \to S^n$ having the property that $p^{-1}\{s_0\}=\partial I^n$, that $p$ is otherwise one-to-one (i.e. it is one-to-one over the set $S^n - \{s_0\}$), and such that a subset $V \subset S^n$ is open if, and only if, $p^{-1}(V) \subset I^n$ is open.

Constructing the function $p$ is most easily done in two steps, first constructing a homeomorphism $(I^n,\partial I^n) \mapsto (D^n,\partial D^n)$ where $D^n$ is the closed unit disc in $\mathbb R^n$, and then following that by a quotient map $D^n \mapsto S^n$ which takes $\partial D^n$ to $s_0$ and is otherwise one-to-one.

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