Definition of map to barycentric subdivision?

Question

I'm reading the Rotman's An Introduction to Algebraic Topology, p.247 and I'm confused with some notation.

What is the $\operatorname{Sd} : K \to \operatorname{Sd} K$ ? Here, $\operatorname{Sd}K$ is the barycentric subdivision.

Note that $\operatorname{Vert}(K) \subset \operatorname{Vert (Sd}K)$ (vertex sets). Let $i$ be the inclusion.

I guess that the $\operatorname{Sd}$ means this inclusion. And my question is, is this inclusion a simplicial map? ; i.e., whenever $\{ p_0, \cdots , p_q \}$ spans a simplex of $K$, then $\{i(p_0),\cdots, i(p_q) \}$ spans a simplex of $\operatorname{Sd}K$?

Answer 1

In the context $K$ should be treated as an abstract simplicial complex.Recall its definition on Rotman's An Introduction to Algebraic Topology, Chapter 7, p.141:

Defintion Let $V$ be a finite set. An abstract simplicial complex $K$ is a family of nonempty subsets of $V$, called simplexes, such that

(i) if $v \in V$, then $\{v\} \in K$;(this condition seems redundant)

(ii) if $s \in K$ and $s' \subset s$, then $s' \subset K$

One calls $V$ the vertex set of $K$ and denotes it by $\mathrm{Vert}(K)$; a simplex $s \in K$ having $q+1$ distinct vertices(elements in $V$ are called vertices I think, though the author doesn't mention it explicitly) is called $q$-simplex.

Hence $K$ is formally a set. Its element is also sets.

And recall the defintion for barycentric subdivision of abstract simplicial complex in the same page:

Example 7.11. if $K$ is an abstract simplicial complex, we construct its barycentric subdivision $\mathrm{Sd} K$ as follows(Here $\mathrm{Sd}K$ is also an abstract simplicial complex):

define $\mathrm{Vert}(\mathrm{Sd}K) = \{\text{simplexes} ~ \sigma: \sigma \in K\}$;

define a simplex in $\mathrm{Sd}(K)$ to be a set $\{\sigma_0, \sigma_1, \dots, \sigma_q\}$ with $\sigma_0 < \sigma_1 < \cdots < \sigma_q$ (where $\sigma < \sigma'$ means $\sigma \subsetneq \sigma'$).

Hence the vertices in $\mathrm{Sd}K$ is not vertex in $K$. From the definition it's all the simplexes in $K$. So formally the veretices in $\mathrm{Sd}K$ are sets. The simplexes in $\mathrm{Sd}K$ formally are sets consisting sets.

Since both $K$ and $\mathrm{Sd}K$ are sets. Now we can define $\mathrm{Sd}$: $$ \begin{align} \mathrm{Sd}: K &\to \mathrm{Sd}K \\ \sigma &\mapsto \{s \in K: s \subset \sigma\} \end{align} $$

Maybe an example(taking from John M. Lee Introduction to Topological Manifolds, Example 5.42(c)) is that $V=\{1,2,3\}$, $K = \{\{1\}, \{2\}, \{3\}, \{1,2\}, \{2, 3\}, \{1, 3\}\}$. Then the vertex set of $\mathrm{Sd}{K}$ is $K = \{\{1\}, \{2\}, \{3\}, \{1,2\}, \{2, 3\}, \{1, 3\}\}$ and (I'm not sure if it's right.) $$ \begin{aligned} \mathrm{Sd}K &= \{\{\{1\}\}, \{\{2\}\}, \{\{3\}\}, \{\{1,2\}\}, \{\{2, 3\}\}, \{\{1, 3\}\}\} \\ &\cup \{\{\{1\}, \{1, 2\}\},\{\{1\}, \{1, 3\}\},\{\{2\}, \{1, 2\}\},\{\{2\}, \{2, 3\}\},\{\{3\}, \{2, 3\}\},\{\{3\}, \{1, 3\}\}\} \end{aligned} $$

Take $\sigma = \{1, 2\} \in K$ as an example, $$ \mathrm{Sd}\sigma = \{\{1\}, \{1, 2\}\} $$

I hope this will help. I'm learning algebraic topology too. It's my first time checking content about simplicial complex. If there is anything wrong about the answer please comment. Thanks.

Edit: I think I made a mistake. For the example $\sigma = \{1, 2\}$, the image of it has two choices. Maybe it's not well defined. But ignoring the meaning of $\mathrm{Sd}: C_*(K) \to C_*(\mathrm{Sd}K)$, it seems that $\mathrm{Sd}_\#$ can be defined by adding the subdivision elements together(Formally by the addition of free abelian group).

Answer 2

Rotman's $\operatorname{Sd}: K \to \operatorname{Sd} K$ does not make sense. It should be some simplicial map because he writes $\operatorname{Sd}_\# : C_\#(K) \to C_\#(\operatorname{Sd} K)$, and such induced chain maps are only defined for simplicial maps.

Now look at the example discussed in onriv's answer:

$V=\{1,2,3\}, K = \{\{1\}, \{2\}, \{3\}, \{1,2\}, \{2, 3\}, \{1, 3\}\}$.

This is the abstract version of a triangle (3 vertices, 3 edges):

.

Its bayrycentric subdivision $\operatorname{Sd} K$ has six vertices and six $1$-simplices and it is easy to see that it can be depicted as

The first simplicial homology of both $K$ and $\operatorname{Sd} K$ is $\mathbb Z$ (see Lemma 9.16 and note that $\lvert K \rvert = \lvert \operatorname{Sd} K \rvert \approx S^1$). The image of any simplicial map $\varphi : K \to \operatorname{Sd} K$ can contain only one or two adjacent vertices of $\operatorname{Sd} K$ (note that edges of $K$ must be mapped to vertices or edges or $\operatorname{Sd} K$. Anyway, $\varphi$ factors through a subcomplex $L \subset \operatorname{Sd} K$ such that $\lvert L \rvert$ is contractible, and therefore $\varphi_\#$ cannot induce an isomorphism between simplicial homology groups (which are non-trivial in dimension $1$). Looking at the proof of Lemma 9.16, we see that the whole approach does not make sense.

So what can be done? In fact one can show that any simplicial approximation $\varphi : \operatorname{Sd} K \to K$ to the identity on $\lvert \operatorname{Sd} K \rvert = \lvert K \rvert$ induces an isomorphism on simplicial homology groups which does not depend on the choice of $\varphi$. I do not know whether its inverse is induced by some "natural" chain map $C_\#(K) \to C_\#(\operatorname{Sd} K)$. If it is, then we could denote it by $\operatorname{Sd}_\#$ although it is not induced by a simplicial map $K \to \operatorname{Sd} K$.