Definition of manifolds with boundary

[PhysicsRock]

TL;DR Summary

Why do we define manifolds with boundary differently from the topological definition of the boundary?

In differential geometry, we typically define the boundary $\partial M$ of a manifold $M$ as all $p \in M$ for which there exists a chart $(U,\varphi), p \in U$ such that $\varphi(p) \in \partial\mathbb{H}^n := \{ x \in \mathbb{R}^n : x^n = 0 \}$. Consequently, we also demand that $M$ is locally homeomorphic to $\mathbb{H}^n := \{ x \in \mathbb{R}^n : x^n \geq 0 \}$, instead of $\mathbb{R}^n$ as in the (usually) previously encountered definitions of topological manifolds.

For such topological manifolds, the boundary is typically defined to be the closure of $M$ without it's interior, i.e. $\partial M_{top} = \bar{M} \setminus \mathring{M}$. Perhaps I'm missing something, but theoretically I don't see any restrictions in this definition that would demand that boundary points are to be mapped onto the boundary of $\mathbb{H}^n$.

My question is, why do we make that alteration?

 

[Orodruin]

theoretically I don't see any restrictions in this definition that would demand that boundary points are to be mapped onto the boundary of Hn.

There is a big difference between needing to be mapped like that and there existing a chart where it is.

 

[PhysicsRock]

There is a big difference between needing to be mapped like that and there existing a chart where it is.

There is a big difference between needing to be mapped like that and there existing a chart where it is.

The way I understand it is that if a point $p \in M$ lies within a chart $(U,\varphi)$ and $\varphi(p) \in \partial\mathbb{H}^n$ then $p$ is considered to be a boundary point. The set of all such $p$ is then called the boundary of $M$.

However, what I don't understand is why we alter the definition from that of the boundary of topological spaces, as given here.

 

[Orodruin]

However, what I don't understand is why we alter the definition from that of the boundary of topological spaces, as given here.

It doesn't, really. It should be clear that $\partial \mathbb H$ is the boundary of $\mathbb H$ and the chart is a homeomorphism.