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Let $A := \mathbb{S}^1 \subset \overline{\mathbb{B}^2}$, and let $f : A \hookrightarrow \overline{\mathbb{B}^2}$ be the inclusion map. Consider the adjunction space $\overline{\mathbb{B}^2} \cup_f \overline{\mathbb{B}^2}$ of $\overline{\mathbb{B}^2}$ to $\overline{\mathbb{B}^2}$ along $f$. I wish to prove that $\overline{\mathbb{B}^2} \cup_f \overline{\mathbb{B}^2}$ is homeomorphic to $\mathbb{S}^2$.

Notation: $\mathbb{S}^1$ is the unit circle, $\mathbb{S}^2$ is the unit sphere, $\overline{\mathbb{B}^2}$ is the closed unit ball; the adjunction space is defined as the quotient $$ X \cup_f Y := \dfrac{X \sqcup Y}{a \sim f(a), \quad \forall a \in A \subseteq Y} $$ for the attaching map $f : A \to X$.

My attempt: construct an explicit quotient map to $\mathbb{S}^2$ and conclude by uniqueness of quotients. Specifically, let $X := \overline{\mathbb{B}^2}$ and $Y := \overline{\mathbb{B}^2}$, where $A$ is now sitting in $Y$ and is embedded into $X$ via $f : A \hookrightarrow X$. I identify $X$ and $Y$ with actual closed balls $\{(x,y) \in \mathbb{R}^2 ~|~ x^2 + y^2 \leq 1\}$, and I define a map $$ \varphi : X \sqcup Y \longrightarrow \mathbb{S}^2 $$ $$ (x,y) \in X \longmapsto \left(x,y, + \sqrt{1 - (x^2 + y^2)}\right), $$ $$ (x,y) \in Y \longmapsto \left(x,y, - \sqrt{1 - (x^2 + y^2)}\right). $$ In other words, send one disc to the upper hemisphere, and the other to the lower hemisphere. Clearly $\varphi$ is surjective; it moreover makes the same identifications as $\sim$. It is therefore a quotient map, completing the proof by uniqueness of quotients.

Uniqueness of Quotients: If $\pi_1 : X \to Y_1$ and $\pi_2 : X \to Y_2$ are two quotient maps that make the same identifications (i.e. $\pi_1 (p) = \pi_1 (q) \iff \pi_2(p) = \pi_2(q)$), then there is a homeomorphism between $Y_1$ and $Y_2$.

Question 1: Are my deductions correct?

Question 2: Is there an easier way of doing this, i.e. without introducing explicit co-ordinates and maps?

Note: cannot use any homology theory, homotopy theory, etc..

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What do you mean by the uniqueness of quotients? The universal property? Does this property imply that the two are homeomorphic? In general it is not true. Another way to conclude this proof is to note that your map is a (well-defined) bijective continuous map between a compact space and a Hausdorff space. –  M.B. Jan 22 '12 at 1:18
    
I've added the explanation of the "uniqueness of quotients". It follows almost immediately from the universal property of quotients. –  Rick Jan 22 '12 at 16:08

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up vote 1 down vote accepted

The homemorphism between $Y_1$ and $Y_2$ is not unique (you can easily find other homeomorphisms). However, there exists a unique homeomorphism $\phi: Y_1 \to Y_2$ such that $\phi\circ\pi_1 = \pi_2$. Other than that (nitpicking) your proof is perfectly fine!

I cannot think of a way to do this without writing out explicit maps. Maybe you could have used some relationship between the mapping cone and the reduced suspension to get the homeomorphism immediately. However, the only way I know to prove that the reduced suspension of the $n$-sphere is the $(n+1)$-sphere is by working with coordinates and maps similar to yours.

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Yes, yes, you're absolutely right, of course, I had missed that part. It's fixed now. Many thanks! –  Rick Jan 22 '12 at 23:19

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