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Let $M$ a $n$-manifold whit boundary, i.e., for each $x\in M$, there exist $U_x\subseteq M$ open in the topology of $M$ such that $U_x$ is homeomorphic to $\mathbb{R}^n$ or homeomorphic to $\mathbb{H}^n$, where

$$ \mathbb{H}^n = \{ (x_1,\ldots,x_n) \in \mathbb{R}^n \;:\; x_n \geq 0\}. $$

Denote by $\partial M $ the boundary of $M$, i.e., $\partial M = \{x \in M\;:\;U_x \cong \mathbb{H}^n \}$.

Suppose that $M$ is embedding in a topological space $X$ and denote by $\partial_T M$ the topological boundary of $M$, i.e., $\partial_T M = X \setminus (Ext(M) \cup Int(M))$.

I conjecture that $\partial M \subseteq \partial_T M$. Is it true? This make sense? If yeah, you can give me a good argument? If not, you can show me a counter-example?

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up vote 2 down vote accepted

You probably had some assumptions in mind, about $M$ or $X$ or the embedding, that you didn't state in the question. What you actually wrote allows the possibility that $X=M$, and then $\partial_TM=\varnothing$.

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Think about the circle in the plane. What is its topological boundary? What condition on $M\subset X$ is necessary and sufficient for $M$ not to be the topological boundary, and hence for $\partial M$ to be the topological boundary?

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Suppose $X$ is a manifold. Since the assertion is local at every point in the boundary, it is enough to consider the case $X=\mathbb R^n$. The dimension $n$ is relevant. If $n-1\ge\dim(M)$, then the topological boundary is the whole $M$, and the assertion holds trivially. If $n=\dim(M)$, then $\text{Int}(M)=M\setminus\partial M$ is an open set in $\mathbb R^n$ and $\partial M$ a hypersurface in $\mathbb R^n$. This hypersurface locally splits $\mathbb R^n$ into two components, one in $M$ and the other in $\mathbb R^n\setminus M$. Consequently the hypersurface $\partial M$ is exactly the topological boundary of $M$. Here the assertion holds more to the point.

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