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In defining singular homology,

A singular $n$-simplex is a continuous mapping $\sigma_n$ from the standard $n$-simplex $\Delta^n$ to a topological space $X$. Notationally, one writes $\sigma_n:\Delta^n\to X$. This mapping need not be injective, and there can be non-equivalent singular simplices with the same image in X.

The boundary of $\sigma_n(\Delta^n)$, denoted as $\partial_n\sigma_n(\Delta^n)$, is defined to be the formal sum of the singular $(n-1)$-simplices represented by the restriction of $\sigma$ to the faces of the standard $n$-simplex, with an alternating sign to take orientation into account. (A formal sum is an element of the free abelian group on the simplices. The basis for the group is the infinite set of all possible images of standard simplices. The group operation is "addition" and the sum of image ''$a$'' with image $b$ is usually simply designated $a+b$, but $a+a=2a$ and so on. Every image $a$ has a negative $−a$.) Thus, if we designate the range of $\sigma_n$ by its vertices


corresponding to the vertices $e_k$ of the standard $n$-simplex $\Delta^n$ (which of course does not fully specify the standard simplex image produced by $\sigma_n$), then

$$\partial_n\sigma_n(\Delta^n)=\sum_{k=0}^n(-1)^k [p_0,\cdots,p_{k-1},p_{k+1},\cdots p_n]$$

  1. So is a singular $n$-simplex a map, unlike a simplex, which is a geometric object?

  2. About the last formula and about orientation: why do we have to alternate signs? I guess that is related to orientation, but I can't get why.

  3. About the last two formulas: Is the expresion starting with [ and ending with ] referring to geometrical objects, like simplices?

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Say we have a triangle with vertices $A,B,C$ going counterclockwise. You would expect the boundary to be $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{AB}‌​+\overrightarrow{BC}-\overrightarrow{AC}$. If you don't alternate signs in the given formula, you will end up with $\overrightarrow{BC}+\overrightarrow{AC}+\overrightarrow{AB}$, which isn't right. Draw a picture! – wj32 Jun 15 '13 at 9:01
I now somehow get that part. Thanks. – Differentio Jun 15 '13 at 9:20
@wj32: Actually, the boundary is $\vec{AB}+\vec{BC}-\vec{AC}$, which is not exactly the same as $\vec{AB}+\vec{BC}+\vec{CA}$, since $\vec{CA}\neq-\vec{AC}$. But there is a close relation between a reverse simplex and the negation of the simplex, namely, $\vec{CA}-(-\vec{AC})$ is a boundary, so they are equal at the level of homology. Of course, the picture of the boundary going counterclockwise is still useful, as it helps you remember the corrects signs of the edges. – Stefan Hamcke Jun 15 '13 at 17:15
up vote 1 down vote accepted

Yes, singular $n$-simplices are maps. We are effectively trying to probe our spaces with maps from the $n$-simplex. Note that these are topologically the same as solid spheres, and their boundary is a sphere. Homology is trying to capture the idea of holes in the space, and so this is a natural thing to consider. We want to see if we can throw a sphere into the space (which has a hole) such that we can't fill the interior of the sphere. It's also important to distinguish between the map (how we've thrown it) and its image — the image could be the same, but how we've thrown it could be different.

Given some singular $n$-simplex $\sigma$, we can consider its restriction to each of the $(n-1)$-simplices which make up its boundary. This is what is being referred to in the formula. We are taking the sum of these restrictions (with a sign), and hence getting a formal sum of $(n-1)$-simplices, which is what we want to construct a chain complex.

Finally, the sign is very important for us to get the cancellation that we want. You see this algebraically when you verify that $\partial \circ \partial$ is zero, and as wj32 indicates, you see this geometrically when you consider the low-dimensional examples.

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