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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?

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what is the gluing lemma ? – mercio Aug 17 '12 at 8:06
up vote 1 down vote accepted

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 $S^2$!

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I can see why the adjunction space is $S^2$. That is clear. I just wanted to assemble a formal argument. I am still thinking on it. – Holdsworth88 Aug 18 '12 at 14:33
@Holdsworth88 Try to proceed using the approach I gave you above. You still not to check it's a quotient map and among other things the quotient space is exactly the one obtained by quotiening out by the equivalence generated by the usual $\sim$ for CW - complexes – user38268 Aug 18 '12 at 14:35
The fact it is a quotient space follows from the fact that it is a surjective map from a compact space to a Hausdorff space. That it makes the same identifications follows from the fact that $f \circ \phi_+=\alpha=i_+\circ q_a$ and $f \circ \phi_-=\beta=i_-\circ q_b$. – Holdsworth88 Aug 18 '12 at 14:41
@Holdsworth88 Yes. The reason is suppose $f^{-1}{U}$ is closed in your compact space. Then a closed subset of a compact set is compact and so $f(f^{-1}(U))$ is compact. By surjectivity of $f$ this is just $U$ so that $U$ is compact. From the fact that a compact subset of a Hausdorff space is closed we conclude that $U$ is closed proving that $f$ is a quotient map. You need to say more about the identifications though. Once you've done that, you're done. – user38268 Aug 18 '12 at 14:45
I have edited the notation to make my question and your solution more compatible. – Holdsworth88 Aug 19 '12 at 7:21

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