Let $X=\mathbb{R^2}-\{0\}$ and $Y=S^1 \cup\{[0,1]\times\{0\}\}$. Prove or Disprove that $X$ is homeomorphic to $Y$.

Question

Let $X=\mathbb{R^2}-\{0\}$ and $Y=S^1 \cup\{[0,1]\times\{0\}\}$. Prove or Disprove that $X$ is homeomorphic to $Y$.

My thoughts:

Since X is the punctured plane, in terms of retraction, I believe that $S^1$ is a deformation retact of $X$, Similarly, the space $Y$ is the unit circle with an extra "strip" along the $x- axis$ same as the radius. And so $S^1$ is a deformation retract of $Y$ as well.

As a result, the fundamental groups of both $X$ and $Y$ are isomorphic to the fundamental group of the unit circle $\pi_1(X,x_0) \approx \mathbb{Z} \approx \pi_1(Y,y_0)$ . Does this isomorphism guarantee a homeomorphism between the spaces $X$ and $Y$? if not, Is there a way to appraoch it without involving fundamental groups?

Answer 1

You can disconnect $Y$ by removing two points but if you remove two points to $X$ it is always connected.

Answer 2

$Y$ is compact, but $X$ isn't.

But $X$ and $Y$ are homotopy equivalent. Both have $S^1$ as a deformation retract.

Answer 3

$HINT:$ Deleting the point $(1,0)$ from $Y$ results it disconnected. Can you find such point in $X$?