Prove that $I^2/\partial I^2$ is homeomorphic to $\mathbb{S}^2$ 
Prove that $I^2/ \partial I^2$ is homeomorphic to $\mathbb{S}^2$

This was the idea I came up with to prove the above. I was thinking that I could prove the above by showing that $I^2/ \partial I^2$ was homeomorphic to the one-point compactification of $(0, 1) \times (0, 1)$ which itself is homeomorphic to the one-point compactification of $\mathbb{R}^2$ which is homeomorphic to $\mathbb{S}^2$. 
So if I could prove that  $I^2/ \partial I^2$ was homeomorphic to $\left((0, 1)^2\right)^* = (0, 1) \times (0, 1) \cup \{p\}$ which is the one-point compactification of $(0, 1) \times (0, 1)$, then I'd have proven the result.
My strategy to prove $I^2/ \partial I^2 \cong \left((0, 1)^2\right)^*$ was to


*

*Define a bijective map $f : I^2/ \partial I^2 \to \left((0, 1)^2\right)^*$

*Prove $f$ is continuous

*Since $ I^2/ \partial I^2$ is compact (since it is the quotient of a compact Hausdorff space) and $\left((0, 1)^2\right)^*$ is Hausdorff by the closed map lemma I can conclude that $f$ is a homeomorphism


To that end I defined a map $f : I^2/ \partial I^2 \to \left((0, 1)^2\right)^*$ by $$f\left(\left[(x, y)\right]\right) = \begin{cases}
(x, y) \ \text{ if } (x, y) \in I^2 \setminus \partial I^2\\
p \ \ \ \ \ \  \ \ \text{ if } (x, y) \in  \partial I^2\\
\end{cases}$$
where $p$ is the point added to $(0,1) \times (0, 1)$ in the one point compactification. It's easy to see that $f$ is bijective, since the equivalence classes of points inside $I^2 \setminus \partial I^2 = (0, 1) \times (0, 1)$, which are just singletons are mapped to 'themselves' again, and the equivalence class of points inside the boundary $\partial I^2$  (which is just a single point in the quotient space) is mapped to a single point in the target space.
Now the only part I'm having a bit of difficulty is in showing that $f$ is continuous. The reason for this is because I'm having a bit of trouble finding a workable 'description' of the open sets in the topology of the one point compactification $\left((0, 1)^2\right)^*$. The topology on $\left((0, 1)^2\right)^*$ is given explicitly by $$\mathcal{T} = \{\text{ open subsets of } (0, 1) \times (0, 1) \} \cup \left\{U \subseteq \left((0, 1)^2\right)^* \ | \ \left((0, 1)^2\right)^*\setminus U \text{ is a compact subset of } (0, 1) \times (0, 1) \right\}$$
and I then (using the Heine-Borel theorem) simplified it to 
$$\mathcal{T} = \{\text{ open subsets of } (0, 1) \times (0, 1) \} \cup \left\{U \subseteq \left((0, 1)^2\right)^* \ | \ \left((0, 1)^2\right)^*\setminus U \cup \{p\} \text{ is closed in  } \mathbb{R}^2 \right\}$$
So now if I pick an open $U$ in $\left((0, 1)^2\right)^*$, then either $U$ is open in $(0, 1) \times (0, 1)$ or $\left((0, 1)^2\right)^*\setminus U \cup \{p\} \text{ is closed in  } \mathbb{R}^2 $. So I have two cases to prove to show that $f$ is continuous.
In the first case when $U$ is open in $(0, 1) \times (0, 1)$ then by the quotient topology on $I^2 / \partial I^2$ it follows that $f^{-1}[U]$ is open in  $I^2 / \partial I^2$ if and only if $q^{-1}[f^{-1}[U]]$ is open in $I^2$ and $q^{-1}[f^{-1}[U]] = (f \circ q)^{-1}[U] = U$ which is open in $I^2$, so $f^{-1}[U]$ is open in $I^2 / \partial I^2$ and we are done.
In the second case when $\left((0, 1)^2\right)^*\setminus U \cup \{p\} \text{ is closed in  } \mathbb{R}^2 $, I'm not sure how to show $f^{-1}[U]$ is open in this case.
So I have two questions. Firstly is my approach to the initial problem correct? If so my question then is: How can I show that $f$ is continuous?
 A: One can define a map $I^n \to  S^n$ by requiring that the interior maps to $S^n \setminus \{pt\}$. You can make this formal by taking the inverse of the stereographic projection $s:S^n \setminus \{pt\} \to \mathbb R^n$ and using your favorite homeomorphism $f:\mathrm{Int}(I^n) \to \mathbb R^n$. I recommend an $\tan$ in each co-ordinate. Composing $s^{-1} \circ f$ gives the desired map.
The key now is to see that this map can be extended to the boundary in a continuous fashion. We can be crude about this by defining $\partial D^n \to pt$ and defining the total map piecewise. 
To check continuity, take any neighborhood of $\{pt\}$ and note that the preimage is just an open ring in $D^n$. Doing this explicitly might be a good exercise, but I don't think its so interesting.
Anyhow, since this makes the appropriate identifications, we know by the universal property of the quotient topology/map that this total map factors through the quotient, and the induced map is bijective. Since the domain is compact and the codomain hausdorff, it is a homeomorphism.
A: Actually there is a simpler way to prove this.

Theorem 1: If $A \subseteq \mathbb{R}^n$ is a compact convex subset with nonempty interior, then $A \cong D^n$ and moreover $\partial A \cong \partial D$.
Theorem 2: $D^n /\partial D^n \cong S^n$

Proof: So using this theorem, since $I^n$ is compact (being a finite product of compact sets) and convex we obtain $I^n \cong D^n$ and $\partial I^n \cong \partial D^n$. Now since $D^n /\partial D^n \cong S^n$ it follows that $I^n / \partial I^n \cong S^n$. $\square$
