Proof verification for Rotman Exercise 6.7 (barycentric subdivision when $n=1$) I was working on Exercise 6.7 of Rotman's book on algebraic topology, and would like to know if the following solution is correct. I'm a bit uncertain because my answer feels less "symmetric" than I would expect.
Definition. Let $E$ be a convex set. Then barycentric subdivision is a homomorphism $\textrm{Sd}_n:S_n(E)\to S_n(E)$ defined inductively on generators $\tau:\Delta^n\to E$ as follows:

*

*If $n=0$, then $\textrm Sd_0(\tau)=\tau$;

*If $n>0$, then $\textrm Sd_n(\tau)=\tau(b_n).\textrm{Sd}_{n-1}(\partial\tau)$, where $b_n$ is the barycenter of $\Delta^n$ and the $.$ represents the cone. In particular, if $\sigma:\Delta^n\to X$ is an $n$-simplex, then $b.\sigma$ is

$$b.\sigma(t_0,t_1,\dots,t_{n+1})=\begin{cases}b&\text{if}~t_0=1\\t_0b+(1-t_0)\sigma\left(\frac{t_1}{1-t_0},\dots,\frac{t_{n+1}}{1-t_0}\right)&\text{if}~t_0\ne1\end{cases}.$$
Then for any space $X$, we define $\textrm{Sd}_n:S_n(X)\to S_n(X)$ as $\textrm{Sd}_n(\sigma)=\sigma_\#\textrm{Sd}_n(\delta^n)$, where $\delta^n:\Delta^n\to\Delta^n$ is the identity.
The problem asks me to find $\textrm{Sd}_n(\delta^n)$ and $\textrm{Sd}_n(\sigma)$ for $n=1,2$. I'll just present my solution for $n=1$, and if anyone could just let me know if it looks correct or not, that'd be great.
My solution for $n=1$.
Since $\textrm{Sd}_0$ is the identity, we know that $\textrm{Sd}_1(\delta^1)$ is $b_1.(\partial\delta^1)$.
But $\partial\delta^1=e_1-e_0$, while $b_1=\frac{1}{2}(e_0+e_1)$, so
\begin{align*}
\textrm{Sd}_1(\delta^1)(t)&=
\begin{cases}
\frac12(e_0+e_1)&\text{if}~t=1\\
t\cdot\frac12(e_0+e_1)+(1-t)(e_1-e_0)&\text{if}~t\ne1.
\end{cases}\\
&= \frac t2(e_0+e_1)+(1-t)(e_1-e_0)\\
&=\left(1-\frac t2\right)e_1+\left(\frac{3t}2-1\right)e_0.
\end{align*}
For general $\sigma$, note that
$$\textrm{Sd}_1(\sigma)=\sigma_\#\textrm{Sd}(\delta)^n=\left(1-\frac t2\right)\sigma(e_1)+\left(\frac{3t}2-1\right)\sigma(e_0).$$
As I briefly mentioned in the beginning, I feel like the formula I end up with looks a little bit asymmetric, so I just wanted to double check my answer.
 A: There is a small conceptual error here to do with the notation:

*

*when you write $b_1 = \frac{1}{2} (e_0 + e_1)$, you mean the $0$-chain consisting of a single map $\Delta^0 \to \Delta^1$ with image $\{\frac{1}{2} (e_0 + e_1)\}$, but

*when you write $\partial \delta^1 = e_1 - e_0$, you mean the $0$-chain consisting of the formal difference of two maps $\Delta^0 \to \Delta^1$ with images $\{e_0\}$ and $\{e_1\}$ respectively.

Therefore, the correct context to use the formula for the cone should be $$\operatorname{Sd}_1(\delta^1) = \delta^1(b_1) . \operatorname{Sd}_0(e_1 - e_0) = b_1 . (e_1 - e_0) = (b_1 . e_1) - (b_1 . e_0).$$  Applying that formula, we get:
$$\begin{align}
(b_1 . e_0)(t) = t b_1 + (1-t) e_0, \\
(b_1 . e_1)(t) = t b_1 + (1-t) e_1,
\end{align}$$
so that $\operatorname{Sd}_1(\delta^1)$ is the $1$-chain given by the formal difference of two maps $\Delta^1 \to \Delta^1$, one from one endpoint of the interval to the midpoint and the other from the other endpoint to the midpoint.  This should agree with the intuition you have about barycentric subdivision.
