Find a homeomorphism

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 $$E=\{(x,y) :(x-1)^2 +y^2 =1\} \cup \{(x,y) :(x+1)^2 +y^2 =1\}$$

• Aren't $A$ and $C$ the same set? – Keen-ameteur Mar 13 at 17:36
• @Keen-ameteur Thanks for pointing that out! Just fixed it. – M. Navarro Mar 13 at 17:40
• Do you want an explicit map or a hint? – Keen-ameteur Mar 13 at 17:42
• By $X/B$ I assume you mean collapsing $B$ to a point, no? – Guido A. Mar 13 at 19:19
• @Keen-ameteur An explicit map – M. Navarro Mar 20 at 16:00

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.