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?