## homotopy invariance of singular cohomology

I am working with Graduate Studies in Mathematics, Volume 65: Global Calculus. I try to understand the proof of homotopy invariance of singular cohomology (Theorem 4.4.6).
Let X be a topological space. Define inclusions $\iota_{0,1} \colon X \rightarrow X \times [0,1], \iota_{i}(x)=(x,i)$
It's enough to show that these maps induce cochain homotopic maps on singular cochain complexes of X and $X \times [0,1]$
Let $f_{i} \colon \mathbb{R}^{n+2} \rightarrow \mathbb{R}^{n+2}$ be the linear map which maps the standard basis vector $e_{j}$ to $(e_{j},0) \ \text{if} \ 0 \le j \le i, (e_{j-1},1) \ \text{if} \ i+1 \le j \le n+1$
If we restrict this map to the standard (n+1)-simplex, it maps to the standard-n-simplex. For a singular n-simplex $\sigma$ define $P_{i}(\sigma):= \sigma \times \identity \circ f_{i}|\Delta_{n+1}$
And further for a singular (n+1)-simplex $\alpha$ in $X \times [0,1]$
$P \alpha (\sigma) := \sum (-1)^{i} \alpha ( P_{i}(\sigma))$
It says "an easy verification gives us the formula":
$d P(\alpha) + P d (\alpha) = \iota_{1}^{*}(\alpha) - \iota_{0}^{*}(\alpha)$ where d is the singular coboundary map, that means: $d\alpha (\sigma) := \sum_{i=0}^{n+1} (-1)^{i} \alpha ( \sigma^{i} )$ ( $\sigma^{i}$ is the i-th face of $\sigma$ )

I tried to recalculate and understand this formula but I don't get it. On the left side there are two double sums and I am too stupid to simplify them. I tried to find out what the terms are but with distinction of cases I don't see why in the first sum are the same terms like in the second with other sign.

I tried to find other books about singular cohomology. But the problem is that almost all define singular cohomology using singular homology and everything from there.
But if it's so easy, has nobody ever prooved the formula and written down somewhere? Or is it so trivial that it's not necessarry? Can anyone help me? I am fortlorn.