A question about Hatcher exercise 2.1.23 
I'm trying to solve a problem on barycentric subdivision. The problem deals with any delta complex in general, so I can't find a way to formulate some argument at all...I can't even see how to express any delta complex in general and apply the barycentric subdivision. Could anyone show me how to solve this problem?
 A: I'll first give a brief excursus about the different ways to look at a delta complex

A general Delta complex $X$ can be described combinatorially as a sequence of sets $X_0,X_1,\dots$ together with maps $d_i:X_n\to X_{n-1}$ for any $i\in[n]=\{0<\dots<n\}$ such that 
  $$ d_j d_i = d_i d_{j+1}, \qquad \text{whenever }\; i\le j $$ whenever $j\ge i$. The elements of $X_n$ are called $n$-simplices of $X$, and the maps $d_i$ are the face maps of $X$.
  Note that any monotone injection $f:[k]\to[n]$ can be written uniquely as a composite 
  $$ f=\delta_{i_{n-k}} \dots \delta_{i_1}, \qquad i_1<\dots<i_{n-k} $$
  where $\delta_i:[m-1]\to[m]$ denotes the monotone injection which omits the value $i$, for any $m$. The indices in the decomposition of $f$ are then simply those elements of $[n]$ which are not assumed by $f$. Such an $f$ can be thought of as representing the face of $\Delta^n$ which is spanned by the basis vectors $e_{f(0)},\dots, e_{f(k)}$. If we denote the restriction of $\sigma$ to this face by $B_f(\sigma)$, then we see that 
  $$
B_f B_g (\sigma) = B_{gf}(\sigma) \quad \text{for any monotone injections $f:[k]\to[l]$ and $g:[l]\to[n]$ }
$$
  This property is actually equivalent to the rule $(1)$, where $d_i=B_{\delta_i}$.

A delta complex $X$ has a realization $|X|$ which is constructed as follows: For any $n$-simplex $\sigma$ take a standard-$n$-simplex $\Delta^n_\sigma$. Now identify its $i$-th subsimplex with the standard-$(n-1)$-simplex $\Delta^{n-1}$ corresponding to the face $d_i\sigma$ using the linear inclusion $\Delta^{n-1} \to \Delta^n$ which preserves the ordering of the vertices. For example, you may have the delta complex $X_0=\{v\} \leftleftarrows X_1=\{e\}$, where the two arrows represent $d_0$ and $d_1$. Then you have a line $\Delta^1$ and a point $\Delta^0$, and both endpoints of the line are glued to that point, so we get a circle.  
The barycentric subdivision $\mathrm{Bd}(X)$ has as $n$-simplices all sequences 
$$
(\sigma, f_1, \dots, f_n), \quad \sigma\in X_m,\, f_i:[k_i]\hookrightarrow[k_{i-1}],\, 0\le k_n<\dots<k_{0}=m
$$
This can be thought of as taking a simplex $\sigma$ and then choosing faces of $\sigma$ such that each face belongs to the previously chosen face. The face map $d_i$ sends $(\sigma, f_n, \dots, f_1)$ to 
$$
\begin{cases}
(B_{f_1}\sigma, f_2, \dots, f_n)  &\text{if } i=0 \\
(\sigma, f_1, \dots, f_{i-1}, f_{i}f_{i+1},\dots, f_n)  &\text{if } 0<i \\
(\sigma,f_1,\dots, f_{n-1})  &\text{if } i=n
\end{cases}
$$
Now can you show that if $X$ has no self-identifications (meaning $B_f\sigma\ne B_g\sigma$ whenever $f\ne g$), then this $\mathrm{Bd}(X)$ produces a simplicial complex?
