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I'm trying to understand the boundary map in singular homology:

  1. What does it mean for a $n$-chain to have no boundary?

Here is my understanding: By definition of singular homology, any $n$-chain is just a finite sum $\Sigma_in_i\sigma_i$ where $\sigma_i$ is a singular $n$-simplex, i.e. a continuous map from $n$-simplex to the space $X$. When we take the boundary map, we are looking at a sum (with signs) of maps on the faces of the $n$-simplex. Those elements in the sum should cancel out in pairs. And any two elements cancel each other out iff they are defined on the same face and their map only differ by a sign.

  1. Can we equivalently describe an element in homology class by a continuous map from a simplicial complex without boundary?

If the above is true, then for an element $A$ in integral homology class of $X$, the maps (with orienation induced from the face, denoted by a plus or minus sign) on the faces cancel out in pairs. Using this pairing relation, we can glue the simplices along their faces(with orientation indicated by their sign in the boundary sum) so that each face lies in the intersection of two simplices. By this way, we have a simplicial complex $A_n$ of dimension $n$ without boundary. Then this homology class can be represented by a continous map from $A_n$ to $X$. Actually, this not only holds for an $n$-chain without boundary, for any $n$-chain, we can use the cancelling pair to glue their faces to get a simplicial complex.

Is the above argument correct? Any comment is appreciated.

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Basically that is correct, although to call this a "simplicial complex without boundary" is kind of meaningless. There is no such object.

There is a terminology for this object, though, it's called a pseudo-manifold. The idea is that in this quotient simplicial complex, for each point $x$ in the interior of an $n$-simplex or an $n-1$ simplex, there exists an open subset $U$ of the simplicial complex such that $x \in U$ and $U$ is homeomorphic to an open subset of $\mathbb R^n$. So, such points $x$ form an open subset of $X$ which is an oriented $n$-dimensional manifold.

What about the points in faces of dimension $n-2$ or less? Well, there's pretty much nothing to be said, one has no control over their local topology, in particular it can probably be a rather arbitrary $n-2$ dimensional simplicial complex.

Anyway, that's exactly what an $n$-dimensional pseudo-Manifold is: a simplicial complex of dimension $n$ such that the union of the interiors of the $n$ simplices and the $n-1$ simplices is an oriented $n$-dimensional manifold.

But the case $n=2$ is somewhat special, because in this case the pseudo-manifold is actually a manifold: each $0$-simplex also has a neighborhood homeomorphic to an open subset of $\mathbb R^2$. And then yes, in the case $n=2$ you get a 2-dimensional manifold without boundary.

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  • $\begingroup$ Thanks for the clarification! This also answers another question I asked actually (I also saw your comment there but I feel like I need to first get this one straight then I can understand the whole thing). Thank you so much! $\endgroup$
    – Lamda8
    Apr 26 at 3:36
  • $\begingroup$ Glad to be of help. $\endgroup$
    – Lee Mosher
    Apr 26 at 14:37

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