If $f$ and $g$ are homotopic, then $H_n(f) = H_n(g)$ proof question From Rotman's Algebraic Topology:

Theorem : If $f$ and $g$ are homotopic, then $H_n(f) = H_n(g)$.
It suffices to construct homomorphisms $P_n^X: S_n(X) \rightarrow S_{n+1}(X \times I)$ with $\lambda_{1 \text{#}}^X - \lambda_{0 \text{#}}^X = \partial_{n+1} P_n^X + P^X_{n-1}\partial_n$.
Where:
$(1.)$ $\lambda_i^X : X \rightarrow X \times I$ by $x \mapsto (x,i)$
$(2.)$ $\lambda_{i \text{#}}^X: S_n(X) \rightarrow S_{n}(X \times I )$ by $\sigma \mapsto \lambda_i^X \sigma$
$(3.)$ $\partial \sigma = \sum_{i=0}^n (-1)^{i} \epsilon_i^n$
$(4.)$ $\epsilon_i^n$ is defned as the affine map $\epsilon_i^n : \Delta^{n-1} \rightarrow \Delta^{n}$ taking vertices $\{e_0, e_1, \dots, e_{n-1}\}$ to $\{e_0, \dots, \hat e_i, \dots, e_n\}$ preserving the displayed orderings: $e_i^n: (t_0, \dots, t_{n-1}) \mapsto (t_0, \dots, t_{i-1}, 0, t_{i}, \dots, t_{n-1})$.

And the first part of the proof where I have my question:

Define $P_0^X(\sigma) : \Delta^1 \rightarrow X \times I$ by $t \mapsto (\sigma(\epsilon_0), t)$, and thendefine $P_0(X) : S_0(X) \rightarrow S_1(X \times I)$ by extending by linearity.
Then evaluating $\partial_1 P_0^X \sigma$, we have $\partial_1 P_0^X \sigma = (\sigma(e_0), 1) - (\sigma(e_0), 0)$.

My question is:
How is $\partial_1 P_0^X \sigma = (\sigma(e_0), 1) - (\sigma(e_0), 0)$?
By definition: $\partial_1 P_0^X \equiv \sum_{i=0}^{1} (-1)^i P_0^X \sigma \epsilon_i^1 \equiv P_0^X \sigma ((0, 1)) -  P_0^X \sigma ((1, 0))$ since $\epsilon_0^1 : \{1\} \mapsto (0, 1)$ and $\epsilon_1^1 : \{1\} \mapsto (1,0)$.
And since $\Delta^1 \equiv \{(t_0, t_1) \in \Bbb R^2 : t_0 + t_1 = 1, t_i \ge 0\} = \{t_0 \in \Bbb R : 0 \le t_0 \le 1\}$ (since $t_1 = 1 - t_0$), shouldn't it equal $\partial_1 P_0^X \sigma = (\sigma(e_0), 0) - (\sigma(e_0), 1)$?
 A: Let's examine in detail what's going on in the proof.
First you start with a function $\sigma: \Delta^0 \rightarrow X$. This function maps $e_0$, which is the only point in $\Delta^0$, to a single point in $X$, namely $\sigma(e_0)$.
You then define $P_0^X(\sigma) : \Delta^1 \rightarrow X \times I$ by $t \mapsto (\sigma(e_0), t)$. What this function is doing is taking a $t$ in the one dimensional simplex $\Delta^1$ (which is homeomorphic to the unit interval $[0,1]$) and mapping it to the point in $X \times I$ with first coordinate a fixed point (the image of the only point $e_0$ in $\Delta^0$ under $\sigma$) and second coordinate $t$. 
So $P_0^X(\sigma)$ is a function with first component the constant map $t \mapsto \sigma(e_0)$ and second component the identity $\operatorname{Id}_{\Delta^1}$
Finally:
$$\partial_1 P_0^X = \sum_{i=0}^{1} (-1)^i P_0^X \sigma \epsilon_i^1 = P_0^X \sigma ((0, 1)) -  P_0^X \sigma ((1, 0))= \\ (\sigma(e_0),\operatorname{Id}_{\Delta^1})((0,1))-(\sigma(e_0),\operatorname{Id}_{\Delta^1})((1,0))=(\sigma(e_0),1)-(\sigma(e_0),0)$$
