Proving $N$ is a topological manifold without boundary.

Let $$M$$ be a topological manifold with boundary $$\partial M$$. Let $$N$$ be a quotient of the topological space $$\{0,1\}\times M$$, two copies of $$M$$, with respect to the relation $$(x,0)\sim (x,1)$$ for all $$x\in\partial M$$. Then $$N$$ is a topological manifold without boundary.

Similar question I found is this one: Poincare duality

1 Answer

Hint: think about the plane as the union of the (closed) upper half-plane $$U$$ and the (closed) lower half-plane $$L$$ (which is a homeomorphic copy of $$U$$). Each of those half-planes is a manifold with boundary $$\partial U = \partial L$$, the $$x$$-axis. Given $$p$$ on the $$x$$-axis, and charts around $$p$$ in $$U$$ and $$L$$, you can glue pieces of those charts together to get a chart around $$p$$ in $$U \cup L$$and so make $$U \cup L$$ into a manifold without boundary (when you include all the charts around points in the interiors of $$U$$ and $$L$$). Now generalise with $$M$$ and $$\partial M$$ in place of $$U$$ and the $$x$$-axis.