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Let $X=A\cup B \cup C$ where $A=\{(x,y) :(x+2)^2 +y^2 =1\}$ and $B=\{x^2+y^2 \leq 1\}$ and $C=\{(x,y) :(x-2)^2 +y^2 =1\}$.

Find a homeomorphism between the quotient space $X/B$ and $E=\{(x,y) :(x-1)^2 +y^2 =1\} \cup \{(x,y) :(x+1)^2 +y^2 =1\}$

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  • $\begingroup$ Aren't $A$ and $C$ the same set? $\endgroup$ – Keen-ameteur Mar 13 at 17:36
  • $\begingroup$ @Keen-ameteur Thanks for pointing that out! Just fixed it. $\endgroup$ – M. Navarro Mar 13 at 17:40
  • $\begingroup$ Do you want an explicit map or a hint? $\endgroup$ – Keen-ameteur Mar 13 at 17:42
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    $\begingroup$ By $X/B$ I assume you mean collapsing $B$ to a point, no? $\endgroup$ – Guido A. Mar 13 at 19:19
  • $\begingroup$ @Keen-ameteur An explicit map $\endgroup$ – M. Navarro Mar 20 at 16:00
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Hint: define a continuous surjection $f : X \to E$. For this, make use of the gluing lemma: the function will be continuous if and only if it is continuous restricted to $A$, $B$ and $C$, as they are closed and their union is the whole of $X$.

Now, show that this map is $B$-compatible, i.e. that if $x,y \in B$ then $f(x) = f(y)$. This proves that $f$ will factor through the quotient, or in other words, that we have a continuous mapping $g : X/B \to E$ defined as $g(x) = [f(x)] \in X/B$.

Finally, prove that $g$ is injective (and thus bijective, as surjectivity comes from the previous factorization) which immediately proves that $g$ is a homeo, as $X/B$ is compact (it is the image via the projection of $X$ which is compact), and $E$ is Hausdorff.

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