How to rigorously prove that $S^1 \wedge S^1 \cong S^2$? Defining $S^1\wedge S^1$ as $S^1\times S^1 \Big/ S^1\vee S^1$ and seeing the torus as a square with opposite sides identified, $S^1\vee S^1$ is just the boundary of the rectangle, so $S^1\wedge S^1$ would be the quotient of the disk $D^2$ with its boundary, that is
$$S^1\wedge S^1 = D^2/\partial D^2 \cong S^2$$
This explanation is okay to me, but I'd like to see a complete proof of this fact, and I don't know where to start. If it's possible I would like some hint on how to proceed rather than a whole answer. Thank you in advance.
 A: One way to make this feel more rigorous is to consider $S^1\times S^1$ as a CW-complex, and play with the quotient spaces to make everything come out right. This won't be the nicest way to go about it, but there will be more gritty details which can be illuminating. Since you don't want a complete answer, I'll give an outline with the details hidden in a spoiler.
First, construct the $1$-skeleton. This is homeomorphic to $S^1\vee S^1$.

 Let $X^0 = \{e_0\}$ and $D_1,D_2\cong\mathbb{D}^1$. Define attaching maps to be the constant maps $f_i:\partial D_i\to X^0$. For $Y_1:= X^0\bigsqcup D_i$, form the quotient space $X^1 = Y_1/(x\sim f_i(x))$ for $x\in\partial D_i$. 

Then attach a single disk  $D$ to create a complex homeomorphic to $S^1\times S^1$.

 Now let $D\cong\mathbb{D}^2$ and define an attaching map $f:\partial D \to X^1$. [For brevity's sake and my own, I'm going to omit the details of this particular attaching map. However, you should be able to construct this yourself.] Then for $Y_2 := X^1\bigsqcup D$, we form the quotient space $X^2 = Y_2/(x\sim f(x))$ for $x\in\partial D$. Let $p$ be the quotient map $p:Y_2\to X^2$. 

Consider the quotient of the 2-skeleton by the 1-skeleton.

 Now we consider the quotient space $X^2/X^1$ with quotient map $q$, and let $A$ denote the equivalence class $[X^1]$. To determine what this is, we follow the chain of quotient maps. For any $x\in(\partial D \cup X^1)$, the image $p(x)\in X^1$, and so $(q\circ p)(x)\in A$. Therefore $\partial D\mapsto A$. For any $y\in D^{\circ}$, by definition of our maps, $(q\circ p)(y) = y$.

Note where $\partial D$ and $D^{\circ}$ are sent. Prove this space is $S^2$.

 Then $r= (q\circ p)$ is a quotient map such that $r|_{D^{\circ}} = id_{D^{\circ}}$ and $r(\partial D)$ is a single point, say $b$. Thus we get a homeomorphism between $X^2/X^1$ and $D/\partial D$ by sending $A\mapsto b$ and acting as the identity everywhere else. Since $D/\partial D\cong S^2$, we're done.

I'm not confident all the details here are correct, so feedback is welcome.
A: one can also use the universal property of quotients (and their uniqueness) to make this work in a slightly more rigorous way. This is a matter of taste, since I view your original argument as perfectly correct.
We see that $(S^1,\{pt\}) \cup (pt,S^1)$ indeed gives the boundary of $I^2$, where we view $T^2$ as the usual identification on the boundary of the unit square. Hence, we see that the composition of quotient maps $f:I^2 \to T^2 \to S^1 \wedge S^1$ necessarily makes the identification $q:I^2 \to I^2/\partial I^2$, so it must factor through $q$ as a map $\tilde{f}$. 
However, note that outside of $\partial I^2$, $f$ is injective, since no identification occurs, and hence $\tilde{f}$ is in fact a bijection since quotients are also surjective. But now, a continuous bijection between a compact and Hausdorff space is necessarily a homeomorphism.
