I just want to ask if there is any deeper motivation or clear geometric "sense" behind the barycentric subdivision. Some friend asked me about this a few months ago, looking back the section at Hatcher, I still feel quite confused. I remember one friend told me combinatorically one can do this from posets back to posets, but this does not give me any way to "understand" it properly. In some books (Bredon, for example), the author use excision property as one of the axioms, I'm wondering "where they came from, why they make any sense?".
[Math] Deeper meanings of barycentric subdivision
at.algebraic-topology
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Lickorish and Martin constructed, for each $r$, a triangulation of the $3$-ball whose $r$th barycentric subdivision collapses, but $(r-1)$th doesn't. The basic idea, going back to Furch and Bing, is to triangulate a cube with a knotted hole, where the missing knot has bridge index $2^r+1$, and then fill back a small part of the hole - so that topologically, no hole remains, but the $3$-ball now contains a knot triangulated by a single edge.
Added later: Kearton and Lickorish also constructed triangulations of the $n$-ball, $n\ge 3$, whose $r$th barycentric subdivision is not collapsible. On the other hand, every triangulation of a ball becomes collapsible after some number of barycentric subdivisions, according to a recent preprint by Adiprassito and Benedetti (see their Corollary 3.5).
I believe $\mathcal D$ is not acyclic, unless $X$ is a discrete space, and therefore $S(X)/\mathcal D$ is not quasi-isomorphic to $S(X)$.
In fact, assuming that I did not make a mistake in the claim below, we have the following description of $\mathcal D$: it is zero in degree zero, and is isomorphic to $S(X)$ in positive degrees. It follows that $H_0(\mathcal D)=0$, $H_1(\mathcal D)$ is the huge group $S_1(X)/\partial S_2(X)$, and for $i>1$ $H_i(\mathcal D)\cong H_i(X)$. However, the inclusion $\mathcal D\hookrightarrow S(X)$ induces zero in homology.
It is obvious that $D_0$ is the zero group. For higher degrees, we have the following claim
Claim: for all $n>0$, the operator $T\mapsto T-sd(T)$ defines an injective homomorphism $S_n(X)\to S_n(X)$. It follows that $D_n\cong S_n(X)$ for all $n>0$.
Sketch of proof (hope it is correct, I did not check every detail): Define a partial ordering on the set of singular $n$-simplices of $X$. Given two singular simplices $\sigma, \tau\colon \Delta^n \to X$, we say that $\sigma\le \tau$ if $\sigma=\tau$, or there exists a contracting linear map $h\colon \Delta^n\to \Delta^n$ such that $\sigma=\tau \circ h$. By contracting we mean that there exists a $0\le r<1$ such that $\mbox{dist}(h(x),h(y))\le r\cdot \mbox{dist}(x, y)$. It is not hard to check that this is a partial order. Moreover $\sigma=\sigma\circ h$ for some contracting map only if $\sigma$ is constant.
Now let $0\ne T\in S_n(X)$. We can write $T=\Sigma_{i=1}^m n_i \sigma_i$, where $\sigma_1, \ldots, \sigma_m$ are pairwise distinct singular simplices, and $n_i$ are non-zero integers. If all $\sigma_i$ are constant, then $T-sd(T)=T\ne 0$. Otherwise, among the $\sigma_i$s there exists a non-constant simplex $\sigma_{i_0}$ that is maximal with respect to the order that we defined. The singular chain $T-sd(T)$ is a linear combination of $\sigma_i$s and simplices that are smaller than $\sigma_i$. It is clear that there is no summand that can cancel $\sigma_{i_0}$, so $T-sd(T)\ne 0$.
Best Answer
There can be many reasons for subdividing simplices, barycentrically or otherwise.
For a simplicial complex (triangulated space) there are the simplicial homology groups. These are known to be isomorphic to the singular homology groups, therefore (1) invariant under homeomorphism, and in particular (2) invariant under (not necessarily barycentric) subdivision. Before the invention of singular homology, I believe that (1) was unknown. Fact (2) was a key part of the theory. Subdivision is important simply because even if your space is made out of simplices you will sometimes care about subsets which are only unions of simplices after you cut the space up finer. In simplicial homology, excision is an easy algebraic fact, stemming from the fact that when a complex is a union of two subcomplexes then every simplex is in one or the other (or both).
In singular theory, as you know, invariance under homeomorphism is a triviality but excision requires some work. The point is that when a space is a union of two open sets then (bad news) not every singular simplex is in one or the other but (good news) simplices can be systematically replaced by combinations of smaller simplices to show that this does not matter. This is where subdivision is used, and there is no reason it has to be barycentric. It's like with the fundamental group: you might explore a space by using maps of a standard unit interval into it, but in proving the Seifert-Van-Kampen Theorem you might want to subdivide that interval into little pieces.
Barycentric subdivision also rises in PL (piecewise linear) topology in one other specific technical way that has nothing much to do with homology: regular neighborhoods. In a finite simplicial complex $K$, the smallest neighborhood of a given subcomplex $L$ that is itself a subcomplex does not in general have $L$ as a deformation retract, but this becomes true if you first barycentrically subdivide twice.
And in the interplay between categories and simplicial constructions barycentric subdivision turns up in various ways.
ADDED in response to Hatcher's answer and its comment thread:
Yes, there is a way of extending to all $n$ the pattern that begins: cut a segment in half, cut a triangle into four equal pieces using midpoints of edges ... It is sometimes called "edgewise subdivision", I believe. It may be realized for simplicial sets as follows: A simplicial set is a functor $\Delta^{op}\to Set$ where $\Delta$ is the category of standard nonempty ordered finite sets; its subdivision is obtained by composing with (the opposite of) the functor $\Delta\to\Delta$ which takes an ordered set to two copies of that set laid end to end. This leaves the realization unchanged. Applied to a standard $n$-simplex, it gives a certain subdivision with $2^n$ pieces. If $n>2$ then the pieces are not all the same shape. If $n=3$ you get a tetrahedron cut into four scaled-down models of itself sitting in the corners and four more whose union is an octahedron; these four all share an edge, the only internal edge that there is. It's not immediately clear to me what diameter estimate is available for the pieces.
This can be generalized so that you now cut an edge into $k$ equal pieces and a triangle into $k^2$ congruent pieces (almost half of which are upside down) and in general cut an $n$-simplex into $k^n$ pieces. This $k$-fold edgewise subdivision plays a role in the area of cyclic homology and related things: when a simplicial set $X$ has the kind of extra structure that makes it a cyclic set (a suitable action of a cyclic group of order $m$ on the set $X_{m-1}$ for all $m>0$) then its realization has an action of the circle group, and to make the action of the subgroup of order $k$ appear as a simplicial action you can do the $k$-fold edgewise subdivision described above.
There is also another edgewise subdivision. In this one the $1$-simplex is cut in half as before and the $2$-simplex is cut into four pieces in the following way: join the middle vertex to the midpoint of the opposite side, and join that midpoint to the midpoints of both of the other sides. This construction corresponds to the functor $\Delta\to\Delta$ that takes an ordered set to two copies of the same laid end to end but with the order reversed in one copy.
See also my answer to the recent MO question "Endofunctors of the Simplex Category".
The second edgewise subdivision that I described can be used to analyze the relationship between two definitions of algebraic $K$-theory: Quillen's $Q$-construction is essentially a subdivision of Waldhausen's $S$-construction.