Are $A$ and $B$ homeomorphic?

Question

Let $A=\left\{ (x,y,z) \in \mathbb{R}^3 : y=0, \, x^2+z^2<1 \right\}\cup\left\{ (x,y,z) \in \mathbb{R}^3 : x=0, \, y^2+z^2<1 \right\}$ be a subspace of the Euclidean space $\mathbb{R}^3$. I was wondering if A is homeomorphic to $B=\left\{ (x,y) \in \mathbb{R}^2 : x^2+y^2<1 \right\}$ and how to prove or disprove that. Thanks for your answers.

Answer 1

One way is to proceed as you would to prove $\mathbb{R}^2$ and $\mathbb{R}^3$ are not homeomorphic.

$A'=A\backslash \{(0,0,0)\}$ deformation retracts onto the union of two circles that intersect at two points. This is a graph with free fundamental group $F_3$ on $3$ generators. Hence $\pi_1(A')\cong F_3$

If $f:A\to B$ was a homeomorphism, let $b_0=f(0,0,0)$ and $B'=B\backslash \{b_0\}$ so that the restriction $f':A'\to B'$ is a homeomorphism. Now $f'$ must induce an isomorphism on $\pi_1$ but $\pi_1(B')\cong\mathbb{Z}$; a contradiction.

Answer 2

No, A is two intersecting disks, B is one disk. 

Every simply connected open subset of B is homeomorphic to B.
Not so for A.