# Proving a quotient space is a topological manifold

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.

• What do you mean by non separated? Non Hausdorff? – Paul Frost Apr 5 '19 at 13:03
• yes, non Hausdorff – 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.
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 $$p(x,i) \in p(U_i) = V_i$$.