What am I missunderstanding for the definition of the $i$-th face map of an $n$-simplex?

Question

From Rotman's Algebraic Topology:

Let $X$ be convex and let $\gamma = \sum m_i \sigma_i \in S_n(X)$. Where $S_n(x)$ is the free abelian group with basis all singular $n$-simplexes in $X$.

Prove that if $b \in X$, then $\partial(b . \gamma) = \{\gamma - b.\partial \gamma \text{ if } n \gt 0, (\sum m_i)b - \gamma \text{ if } n =0 \}$.

Let $\sigma$ be a $0$-simplex. $b.\sigma: \Delta^1 \rightarrow X$ is defined as $b.\sigma(t_0, t_1) = \{b \text{ if } t_0 = 1, t_0b + (1-t_0) \sigma(\frac{t_1}{1-1_0}) \text{ if } t_0 \neq 1\}$ or $b.\sigma(t_0) = t_0b + (1-t_0) \sigma(1)$ since $t_0 + t_1 = 1$.

$\partial(b.\gamma) = \sum m_i \partial (b.\sigma_i) = \sum m_i(b - x_i) = (\sum m_i)b - \gamma$.

In the last line, how is $\sum m_i \partial (b.\sigma_i) = \sum m_i(b - x_i)$?

By definition of $\partial \sigma = \sum_{j=0}^n (-1)^j\sigma \epsilon_j^n$, where $\epsilon_j^n$ is the $j$th face map, $$\partial (b.\sigma_i) = b.\sigma_i \epsilon_0^1 - b.\sigma_i\epsilon_1^1 $$ $$= b.\sigma_i((0, 1)) - b.\sigma_i((1,0)) = b.\sigma_i(0) - b.\sigma_i(1) = \sigma_i(1) - b = x_i - b$$

What definition am I misunderstanding or has the author made a mistake in the statement of the theorem?

Answer 1

When proving Theorem 4.14 (i) in page 70, the author says:

[...] Note that $\partial_1 \sigma_i = \sigma_i(e_1)-\sigma_i(e_0) = x_i-x$ (We have identified $I=[0,1]$ with $\Delta^1=[e_0,e_1]$).

What the last sentence means is that he has considered an homeomorphism between $I=[0,1]\subset \mathbb{R}$ and $\Delta^1$ (which is the segment in $\mathbb{R}^2$ connecting the points $e_0=(1,0) $ and $e_1=(0,1)$) and is identifying the points of $\Delta^1$ with their homeomorphic image. This in particular means that he's identifying the corresponding endpoints of both segments, i.e.:

$$ 0 \sim e_0 = (1,0) \\ 1 \sim e_1 = (0,1)$$

Let's consider the equality:

$\partial(b \cdot \gamma) = \sum m_i \partial (b \cdot \sigma_i) = \sum m_i \big(b \cdot \sigma_i \epsilon_0^1 - b \cdot \sigma_i\epsilon_1^1 \big)$

Now the functions $b \cdot \sigma_i \epsilon_0^1$ and $b \cdot \sigma_i \epsilon_1^1$ do the following:

\begin{array}{cccl} b \cdot \sigma_i \circ \epsilon_0^1: &\Delta^0 & \stackrel{\epsilon_0^1}{\longrightarrow} & \Delta^1 & \stackrel{h}{\longrightarrow} & I & \stackrel{b \cdot \sigma_i}{\longrightarrow} & X \\\ &1 &\longmapsto & (0,1) & \stackrel{\sim}{\longmapsto} & 1 & \longmapsto & 1 \cdot b + 0 \cdot x_i \end{array}

\begin{array}{cccl} b \cdot \sigma_i \circ \epsilon_1^1: &\Delta^0 & \stackrel{\epsilon_1^1}{\longrightarrow} & \Delta^1 & \stackrel{h}{\longrightarrow} & I & \stackrel{b \cdot \sigma_i}{\longrightarrow} & X \\\ &1 &\longmapsto & (1,0) & \stackrel{\sim}{\longmapsto} & 0 & \longmapsto & 0 \cdot b + 1 \cdot x_i \end{array}

(Note that $1\in \Delta^0$ and $1\in I$ are not the same object: the former is the only point of the 0-simplex and the latter is a point of the unit interval $[0,1] \subset \mathbb{R}$)

Finally: $$\partial(b \cdot \gamma) = \sum m_i \big(b \cdot \sigma_i \epsilon_0^1 - b \cdot \sigma_i\epsilon_1^1 \big)= \sum m_i \big(b -x_i)$$