Let $Y=\Bbb R^2\times\{0,1\}/\sim $ where $\sim$ is the equivalence relation generated by $(x,0)\sim(x,1)\ \forall x\in \Bbb R^2 \setminus\{0\}$.

Question: Show that $Y$ equipped with the quotient topology is a topological manifold of dimension 2, connected and non separated.

To show that it's a manifold of dimension 2 we can construct a map $\psi:U\rightarrow \Bbb R^2$ where $U$ is an open subset of $Y$, and show that it is a homeomorphism which I think is fairly striaghtforward if we take the map $\psi(\overline{(x,y)}) = x$

However I am not sure how to start with showing that it is connected and non separated.

  • $\begingroup$ What do you mean by non separated? Non Hausdorff? $\endgroup$ – Paul Frost Apr 5 '19 at 13:03
  • $\begingroup$ yes, non Hausdorff $\endgroup$ – user520830 Apr 5 '19 at 13:11

$Y$ is a two-dimensional version of the line with two origins. There are many posts in this forum dealing with the latter, but as far as I can see there are no proofs that it is connected and non-Hausdorff. So let us do it here. Let $p : \Bbb R^2\times\{0,1\} \to Y$ denote the quotient map.

1) Y is path connected (which implies that it is connected).

We have $Y = Y_0 \cup Y_1$ with $Y_i = p(\Bbb R^2\times\{i\})$. Both $Y_0, Y_1$s are path connected (they are continuous images of path connected spaces). Since $Y_0 \cap Y_1 \ne \emptyset$, their union $Y$ is path connected.

2) $Y$ is not Hausdorff.

The two points $y_i = p(0,i) \in Y, i = 0,1$, are distinct. Let $V_i$ be open neighborhoods of $y_i$ in $Y$. Then $U_i = p^{-1}(V_i)$ are open neigborhoods of $x_i =(0,i)$ in $\Bbb R^2\times\{0,1\}$. Choose open neigborhoods $W_i$ of $0$ in $\Bbb R^2$ such that $W_i \times \{ i \} \subset U_i$ and choose $x \in W_0 \cap W_1, x \ne 0$. Then $p(x,0) = p(x,1)$. This shows that $V_0 \cap V_1 \ne \emptyset$ because $p(x,i) \in p(U_i) = V_i$.


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