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I am getting confused with the concept of boundary. So I would like to see what a boundary is by using examples. So does the torus have a boundary?

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    $\begingroup$ Boundary in what sense? Boundary as a manifold? No. Relative boundary within itself? $\endgroup$ – Alex Youcis Mar 30 '13 at 7:41
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    $\begingroup$ Just so we're all on the same page: By "torus," you mean the 2-dimensional (hollow) surface that lives in $\mathbb{R}^3$, right? Because this is different from the "solid torus," which is a 3-dimensional (solid) volume that also lives in $\mathbb{R}^3$. $\endgroup$ – Jesse Madnick Mar 30 '13 at 9:17
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    $\begingroup$ Still confused? I would guess that you are :-) $\endgroup$ – bubba Mar 30 '13 at 11:46
  • $\begingroup$ Now grasped the concept. So solid cylinder has no boundary then, right? (As in manifold boundary) $\endgroup$ – yous Mar 30 '13 at 18:01
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One has distinguish the boundary of a manifold and the boundary of subsets of a topological space (which may be confusing, since a manifold is a topological space with further structure).

A manifold with boundary $M$ is often defined as a topological Hausdorff space locally homeomorphic to the upper half space $\mathbb{H}^n:=\{(x^1,\dots,x^n)\in\mathbb{R}^n:x^n\geq 0\}$; in contrast to a manifold without boundary, which locally is homeomorphic to open subsets of $\mathbb{R}^n$. (Its boundary $\partial\mathbb{H}^n$ are all points $x\in\mathbb{H}^n$ satisfying $x^n=0$.Then, in a manifold sense, the boundary $\partial M$ of $M$ is the inverse image of $\partial H^m$ under some chart.)

So, if one wants to know whether a manifold has boundary or not you have to look at its atlas, i.e. the given structure.

(For details on this subject I recommend Lee: Introduction To Smooth Manifolds)

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    $\begingroup$ +1: nice, short answer. It transpires from what you say, but a manifold is locally homeomorphic to $\Bbb R^n$ (and similarly for a manifold with boundary), i.e. every point in it admits an open neighbourhood homeomorphic to $\Bbb R^n$. $\endgroup$ – A.P. Mar 30 '13 at 9:30
  • $\begingroup$ You are right, I tend to forget mentioning. $\endgroup$ – gofvonx Mar 30 '13 at 12:58
  • $\begingroup$ @A.P.: What is the neighbourhood of $0$ homeomorphic to $\mathbb{R}$ in the upper half-space $\mathbb{H}^1 = [0,\infty)$? $\endgroup$ – Tara B Mar 30 '13 at 14:51
  • $\begingroup$ @TaraB In my comment I said specifically and similarly for a manifold with boundary, meaning that in that case there could be some points with a neighbourhood homeomorphic to $\Bbb H^n$. Sorry if this wasn't clear enough. $\endgroup$ – A.P. Mar 30 '13 at 14:53
  • $\begingroup$ Ah, I see. I'm not sure if it would be clear to everyone what 'similarly' is supposed to mean there. I do see that it is a reasonable meaning, though. But I think it could be slightly misleading. $\endgroup$ – Tara B Mar 30 '13 at 14:55
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Torus does not have a boundary (when viewed as in a differential geometry course).

This may be slightly confusing if you think of torus as the usual "doughnut" hanging in the three dimensional space. However, what is meant by torus is really the surface of this figure, with it's own intrinsic geometric structure. The torus can also be viewed as an abstract manifold, without reference to the three dimensional space, by identifying the opposite sides of a square --- then it no longer feels like it should have a boundary, at least not to me. The important thing is that locally torus looks like a (piece of a) plane.

A manifold with boundary is something that does not look like a plane everywhere, but has places (boundary) where it look locally like a half-plane. Take for instance a circle, together with the boundary $\{(x,y)\in\mathbb{R^2}\ : \ x^2+y^2 \leq 1\}$. It is a manifold with boundary, because at the boundary points like $(0,1)$ it does not look like a plane. However, an open circle $\{(x,y)\in\mathbb{R^2}\ : \ x^2+y^2<1\}$ is a manifold without boundary.

A final note: if you actually wanted torus to be a doughnut, i.e. a $3$-dimensional thing, then it will be a (rather uninteresting) manifold with or without boundary, depending on whether you include the boundary.

And one more thing: there is also a notion of boundary in topology. This is probably not quite what you mean. It applies to a subset of a topological space (so you can't ask for a boundary of the torus, unless you specify where it sits), and is usually non-empty.

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  • $\begingroup$ Good one. But i couldn't understand your statements i.e. $\hspace{9cm}$ The torus can also be viewed as an abstract manifold, without reference to the three dimensional space, by identifying the opposite sides of a square --- then it no longer feels like it should have a boundary, at least not to me. The important thing is that locally torus looks like a (piece of a) plane. $\hspace{9cm}$ Two things which I didn't understand is (i) by identifying the opposite sides of a square, and (ii) The important thing is that locally torus looks like a (piece of a) plane. $\endgroup$ – L.K. Apr 25 '17 at 9:52
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As illustrated by the above answers, there are several meaning of the term boundary, and the precise notion of the boundary of a subset of a topological space is one of them.

Another is of course the notion of the boundary of an $n $-manifold with boundary. So if we consider a solid disk $D $ in the plane it is a $2 $-manifold with boundary a circle $C$, and it's boundary as a subset of the plane is also the same circle.

The notion of "boundary" has a more confusing history in algebraic topology, where the early writers (e.g. Betti) wanted to talk about "cycles modulo boundaries" but the notions were not clear. Poincaré's famous papers on "Analysis situs" (starting in 1895) cleared up many things by using simplicial decompositions and the notion of $n $-chain as a formal sum of oriented $n $-simplices, so that the boundary of an $n$-simplex had a well defined meaning. This was further clarified by Eilenberg (1944) with the notion of an ordering on the vertices of a simplex, so the that the $i$'th face $\partial_i \sigma$ was well defined and we could then define $$\partial \sigma = \sum _i (-1)^i \partial _i \sigma$$ leading to $\partial \partial=0 $. Hence a cycle was a chain $z $ with $\partial z=0 \,$ and a boundary $b $ was a chain such that $b=\partial c $ for some $c $. Then every boundary was a cycle, as required for the notion of homology.

There is more to be said on this story. But the point I wanted to make is that the notion of "boundary" has a long history and it is not unreasonable to be confused on the intuition.

Nov 19, 2016 See this paper by M. Barr on the relations between oriented and ordered singular theories.

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