# Smooth structure on $M\cup_f N$?

Let $M$ and $N$ be two smooth manifolds with $$\textrm{dim}(M)=\textrm{dim}(N)=n.$$ Let $U\subseteq M$ and $V\subseteq M$ be two open sets and $f:U\longrightarrow V$ a smooth diffeomorphism. Consider the equivalence relation in the disjoint union $M\sqcup N$: $$x\sim \phi(x), \forall x\in U.$$ Consider $M\cup_f N:=(M\sqcup N)/\sim$ with the quotient topology.

Is it true that there is a unique smooth structure on $M\cup_f N$ such that the inclusions $\imath:M\longrightarrow M\cup_f N$ and $\jmath:N\hookrightarrow M\cup_f N$ are embeddings?

• Just drawing pictures, it's not clear to me that $M\cup_f N$ is a manifold at all. For example, if you do this where $M = N = S^1$, don't you get something homeomorphic to the number $8$? – Jason DeVito Nov 13 '14 at 15:24
• Do you require your manifolds to be Hausdorff? Because in general, $M\cup_f N$ will not be. A simple counterexample is obtained by taking $M=N=\mathbb R$, $U=V=\mathbb R\setminus \{0\}$, and $f\colon U\to V$ to be the identity map. Than $M\cup_f N$ is the "line with two origins," which is not Hausdorff. – Jack Lee Nov 13 '14 at 16:35
• Yes @JackLee I suppose both $M$ and $N$ are Hausdorff. – PtF Nov 13 '14 at 17:56
• @JasonDeVito Since I'm not sure about the result I changed my question to "is it true that..".. – PtF Nov 13 '14 at 17:58
• @PtF: Then I think my comment is a counter example to your question. In fact, I think that most of the time $M\cup_f N$ is not a manifold at all. (In fact, I don't know of a single situation, other than $U = V = M = N$ where the $M\cup_f N$ has the structure of a manifold at all.... – Jason DeVito Nov 13 '14 at 18:03

Since your comment made clear that your definition of manifold includes the Hausdorff condition, the answer is no: in general, $M\cup_f N$ will not even be a topological manifold. A simple counterexample is obtained by taking $M=N=\mathbb R$, $U=V=\mathbb R\smallsetminus \{0\}$, and $f\colon U\to V$ to be the identity map. Then $M\cup_f N$ is the "line with two origins," which is not Hausdorff.