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Let $A$ be the open unit disc in $\mathbb{R}^2$ and $B$ be the closed unit disc in $\mathbb{R}^2$.

The toplogical boundary of $A$ and $B$ is $S^1$. This I understand.

The manifold boundary of $A$ is $\emptyset$ and the manifold boundary for $B$ is $S^1$. This I don't understand. Can someone briefly explain?

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In a topological manifold $M$ (possibly with boundary), each point $p \in M$ must have an open neighborhood $p \in U$ which is homeomorphic either to an open set in $\mathbb{R}^n$ or to an open set in $\mathbb{H}^n := \{ (x_1, \ldots, x_n) \in \mathbb{R}^n \, | \, x_n \geq 0 \}$. The set $\partial \mathbb{H} := \{ (x_1, \ldots, x_{n-1}, 0) \, | \, x_1, \ldots, x_{n-1} \in \mathbb{R} \}$ is called the boundary of $\mathbb{H}$.

Let $p \in M$ and let $\varphi \colon U \rightarrow V$ be a homeomorphic between an open neighborhood of $p \in U$ in $M$ and an open neighborhood in $\mathbb{R}^n$ or $\mathbb{H}^n$. If $V \subseteq \mathbb{H}^n$ and $\varphi(p) \in \partial \mathbb{H} \cap V$, we say that $p$ is a boundary point of $M$. If $V \subseteq \mathbb{R}^n$ or $V \subseteq \mathbb{H}^n$ and $\varphi(p) \notin \partial \mathbb{H}^n$, we say that $p$ is an interiour point. It is a non-trivial result that the set of interior points and the set of boundary points is disjoint.

Consider for example the closed unit disk $B = B(0,1) \subseteq \mathbb{R}^2$. Using polar coordinates, you can easily construct a homeomorphism from an open neighborhood of each point $x \in B$ with $|x| = 1$ to an open neighborhood of $\mathbb{H}^2$ mapping $x$ to $\partial \mathbb{H}^2$. For $x \in B$ with $|x| < 1$, the open ball $\{ x \, | \, |x| < 1 \}$ is a neighborhood that is an open subset of $\mathbb{R}^2$. Thus, the boundary of the closed unit disk as a manifold is the same as the topological boundary. Since the open unit disk is already an open subset of $\mathbb{R}^2$, the boundary of it considered as a manifold is empty.

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You need to review your definition of a manifold with boundary v.s. manifold without boundary.

In the case of A, for every point $p \in A$ you can find a neighbourhood $U$ such that $p \in U$ and a diffeomorphism $\phi: U \to V \subset \mathbb{R}^2$ (the natural inclusion), where $V$ is open in $\mathbb{R}^2$.

In the case of B, for every point $p \in S^1 \subset B$ you cannot find a neighborhood $U$ of $p$ such that $U$ is diffeomorphic (even homeomorphic) to some open subset $V$ of $\mathbb{R}^2$, but you can find one diffeomorphic to an open subset of some half-space $\mathbb{R}^2_{\lambda \geq c}$ making it a manifold with boundary.

NOTE: When you say that the topological boundary of $A$ is $S^1$ you are thinking of A as a subset of $\mathbb{R}^2$ and when you think of A as a manifold you think it as the whole space.

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