Showing the Adjunction Space $D^2\cup_{f} D^2$ is Homeomorphic to $S^2$ Let $D$ be the closed unit disk. I am trying to show the adjunction space $D^2\cup_{f} D^2$ given by the identification map $f:S^1\rightarrow D^2$ is homeomorphic to $S^2$. 
The following is my attempt:
By the uniqueness of the quotient space it is enough to show that there is a quotient map $q:D^2\amalg D^2\rightarrow S^2$ making the same identifications as the adjunction space. For the sake of identification, I label the different copies of $D^2$ in $D^2\amalg D^2$ by $a$ and $b$, so from now on I will write $D^2 _a\amalg D^2 _b$. Let $(S^2)^+$ and $(S^2)^-$ denote the upper an lower hemisphere of the sphere respectively. Define the continuous maps $q_a: D^2 _a \rightarrow (S^2)^+$ to be the map that projects a point of the closed unit disk $D _a$ to the point on the sphere above it and $q_b:D^2\rightarrow (S^2)^-$ to be the map that projects a point of the closed unit disk $D^2 _b$ to the point on the sphere below it. Using the gluing lemma, there exists a continous function $q:D^2 _a\amalg D^2 _b \rightarrow S^2$ that restricts to $q_a$ and $q_b$ on their domain and agrees on their areas of intersection, which is empty. To see that $q$ is a quotient map note that it is surjective, $D^2 _a\amalg D^2 _b$ is compact and $S^2$ is Hausdorff. This map makes the same identifications as the adjunction, hence by the uniqueness of the quotient they must be homeomorphic. 
My question is:

Have I used the gluing lemma in the correct way?

Also, there is the implicit question:

If I have not misused the gluing lemma, is this proof correct?

 A: I think I would avoid using the gluing lemma because in there you need the hypothesis:


"If for all $x \in A \cap B$ we have $f(x) = g(x)$ then..."


It is tricky to deal with getting around a statement like that. I am not too sure myself on how to use it here, so I will appeal to the universal property of the disjoint union. However, in the universal property you have maps out of your spaces $X_i$ into one single space so we get around this by composing $q_a$ and $q_b$ with $i_{\pm} : (S^2)^{\pm} \hookrightarrow S^2$ so that $\alpha = i_+ \circ q_a$ and $\beta = i_- \circ q_a$ are maps from (two copies of) $D^2$ to $S^2$. To differentiate between these two copies or rather to emphasize that we have two copies, we write $D^2_+$ and $D^2_-$  Now we are all set: by the universal property of the disjoint union, there exists precisely one continuous function 
$$f : D^2_+ \sqcup D^2_- \to S^1$$
such that $f \circ \phi_+ = \alpha$ and $f \circ \phi_- = \beta$. The $\phi_{\pm}$ are the canonical injections of $D^2_{\pm}$ into $D^2_+ \sqcup D^2_2$. It now remains to check that $f$ is a quotient map.
Edit: I think it is more useful to think graphically why adjunction space is indeed $S^2$. In terms of $CW$ complexes you are attaching a $2$ - cell to $D^2$ where the attaching map is the restriction of the identity map  $id : D^2 \to D^2$ to $S^1$. As I had ravioli for dinner, here are instructions on how to see the homeomorphism.
If you like fresh pasta roll out a thin sheet of it and cut out two circles. On one of them, spread egg yolk on the circumference. Now take your other sheet of pasta and glue it onto the one that you spread egg yolk on. Because we are only gluing the edges, grab the center of each sheet and try to pull one above and the other below without tearing the edges. You now see why the adjuntion space is $S^2$!
A: I see no problem with your proof. Just a small detail that is also mentioned in the other answer: there is no need to restrict the domain of the maps $q_a$, $q_b$ to $(S^2)^+$ and $(S^2)^-$, respectively. The gluing lemma works if you let their domains be $S^2$. This eliminates the need to introduce inclusion maps $i_{\pm}:(S^2)^{\pm}\to S^2$. Also, I'd rather use the gluing lemma than the universal property of the disjoint union, as the former is more general in the sense that allows you to construct continuous maps from spaces that are not disjoint unions. As you said, the hypothesis that $q_a$ and $q_b$ agree on the intersection of their domains is trivially satisfied, so everything looks fine for me.
