Barycentric subdivision preserves geometric realization I have the following definitions:
Definition 1: A simplicial complex $K$ is a family of finite nonempty subsets of a set $V_k$ (the elements of $V_k$ are called vertices) such that:
1) if $v\in V_k$, then $\{v\}\in K$.
2) if $S\in K$ and $\emptyset \neq T\subseteq S$, then $T\in K$.
The elements of $K$ are called simplexes. 
Definition 2: The geometric realization of a simplicial complex $K$ is defined as $|K|= \{\alpha: V_K \to [0,1]  \; | \; \alpha^{-1}((0,1]) \in K , \;\sum_{v\in V_k}  \alpha(v)=1  \}$.
Definition 3: The barycentric subdivision of a simplicial complex $K$ is defined as the simplicial complex $sd(K)$ such that $V_{sd(K)}=K$ and a simplex in $sd(K)$ has the form $\{s_0, s_1, ...,s_q\}$, where $s_i \in K $ and $s_{i-1} \subset s_i $,  ($s_{i-1}$ is a proper subset of $s_i$ ), for all $i$.
I'm asked to show that $|sd(K)|$ is homeomorphic to $|K|$.
I already know that for $S=\{v_0,v_1,...,v_q\} \in K$, $|S|=\{\alpha \in |K| \; | \; \alpha^{-1} ((0,1]) \subseteq S\}$ is homeomorphic to $\Delta^q = \{(t_0,t_1,...,t_q)\in \mathbb{R}^{n+1} \; | \; t_i \geq 0, \sum_{i=0}^{n} t_i =1\}$ (actually, the homeomorphism is an isometry).
Here, the topology we are using is the one induced by the metric $d:|K|$x$|K|\to \mathbb{R}$ such that $d(\alpha, \beta) = \sqrt{\sum_{v\in V_K} (\alpha (v) - \beta (v))^2}$.
Now, I'm trying to construct an homeomorphism from $|K|$ to $|sd(K)|$, but everything I can think of, just doesn't work. I would be very thankful if you could give me a hint on how to construct this homeomorphism, or if you could tell me another way to proceed on this. Thank you so much!
 A: Let us think of a function $\alpha:V_K\to [0,1]$ as a formal sum $\sum_{v\in V_K}\alpha(v)\cdot v$.  The homeomorphism $|sd(K)|\to |K|$ is then given by mapping $S\in V_{sd(K)}=K$ to the formal sum $\sum_{v\in S}\frac{1}{|S|}\cdot v$, and "extending linearly".  In other words, a point $\alpha\in|sd(K)|$ can be written as a formal sum $\sum_{S\in K} \alpha(S)\cdot S$ and then we map this formal sum to $|K|$ by replacing each $S$ with $\sum_{v\in S}\frac{1}{|S|}\cdot v$ to get the formal sum $\sum_{S\in K}\sum_{v\in S}\frac{1}{|S|}\cdot v$ which we can think of as a formal sum of elements of $V_K$.  I'll leave it to you to show that this is a homeomorphism.  (As a hint for showing it is a bijection, given a point of $|K|$, you can figure out what simplex of $|sd(K)|$ it is in the image of by ordering its nonzero coefficients from smallest to biggest.  For instance, a sum like $0.2a+0.3b+0.5c$ for three vertices $a,b,c$ can be split up as $0.6(\frac{1}{3}a+\frac{1}{3}b+\frac{1}{3}c)+0.2(\frac{1}{2}b+\frac{1}{2}c)+0.2c$, to write it as a convex combination of expressions of the form $\sum_{v\in S}\frac{1}{|S|}\cdot v$.)
The intuition here is that for each geometric simplex $|S|$ in $|K|$, we add a new vertex at the barycenter of $S$, the point $\sum_{v\in S}\frac{1}{|S|}\cdot v$ that is the average of all its vertices.  We can then take convex combinations of the barycenters of $|S|$ and its faces to subdivide $|S|$ into many smaller simplices.  For a very simple example, suppose $K$ is just a line segment, with two vertices and an edge connecting them.  Then $sd(K)$ has three vertices: two are just the vertices that $K$ had, but one of them comes from the edge.  We think of this new vertex as being the midpoint of the edge of $K$, which we can connect to the two vertices of $K$ to split the edge into two smaller edges.  So, $|sd(K)|$ is just two line segments joined together, which is homeomorphic to a single line segment $|K|$.
