About the existence of homotopy operator What shown below is a reference from the text Analysis on Manifolds by James Munkres. I point out that the sybols $\Omega$ and $^*$ are used by Munkres to denote a set of differential forms and the pull back map respectively.

Definition
Let be $A$ and $B$ open sets in $\Bbb R^n$ and $\Bbb R^m$ respectively and let be $g$ and $h$ two $C^r$ map defined from $A$ into $B$. So we say taht $g$ and $h$ are differentiably homotpic of class $C^r$ if there exists a $C^r$ mapd $H$ from $A\times[0,1]$ into $B$ such that
$$
H(x,0):=g(x)\,\,\,\text{and}\,\,\,H(x,1)=h(x)
$$
for any $x\in A$. The map $H$ is called a differentiable homotopy between $g$ and $h$.
Theorem
Let be $A$ and $B$ open sets in $\Bbb R^n$ and $\Bbb R^m$, respectively. Let $g,h:A\rightarrow B$ be $C^\infty$ maps that are differentiably homotopic. Then there is a linear transformation
$$
\mathcal P:\Omega^{k+1}(B)\rightarrow\Omega^k(A)
$$
defined for $k>0$ such that for any form $\eta$ of order $k>0$
$$
d(\mathcal P\eta)+\mathcal P(d\eta)=h^*\eta-g^*\eta
$$
while for a form of $f$ of order $0$,
$$
\mathcal P(df)=h^*f-g^*f
$$

So to follow I summarise the proof of the previous theorem but if you like you can read it
here.
Well first of all Munkres consider a very special case. So given a neighborhood $U$ of $A\times [0,1]$ in $\Bbb R^{n+1}$ he defines the map $\alpha,\beta:A\rightarrow U$ given by the equations
$$
\alpha(x):=(x,0)\,\,\,\text{and}\,\,\,\beta(x):=(x,1)
$$
and thus observing that these two maps are differentiably homotopic he proves the existence of a linear operator $P$ from $\Omega^{k+1}(U)$ to $\Omega^k(A)$ havig the desidred properties. Then the general case follows by this. Inded if $g,h:A\rightarrow B$ are differetiably homotopic with respect $H:A\times [0,1]\rightarrow B$ then he defines the operator $\mathcal P$ through the equation
$$
\mathcal P\eta:=P(H^*\eta)
$$
for any $\eta\in\Omega^{k+1}(B)$ because (Munkres himself states what to follow) $H^*\eta$ is a $(k+1)$-form defined in a neighborhood of $A\times[0,1]$ and thus $P(H^*\eta)$ is a $k$-form defined in $A$.
So I do not really understand why the pull back of $H$ with respect $H$ is a $(k+1)$-form defined in $A\times[0,1]$. Indeed if $H$ goes from $A\times[0,1]$ to $B$ then the pull back $H^*$ goes from $\Omega^{k+1}(B)$ to $\Omega^{k+1}(A\times[0,1])$ and thus this is a very big problem: in particular it seems to me that if we consider the restriction $\tilde H$ of $H$ to $A\times[0,1)$ then the pull back $\tilde H^*$ define a $(k+1)$-form on $A\times[0,1)$ of class $C^r$ that can be extended to a $(k+1)$-form of class $C^r$ defined in an open neighborhood of $A\times[0,1)$ but unfortunately this open neighborhood is not generally a open neighborhood of $A\times[0,1]$. So why $H^*\eta$ is a form defined in an OPEN neighborhood $A\times[0,1]$? Could someone explain this to me, please?
 A: I don't understand the relevance of restricting $H$ to $A\times [0,1)$, but I think I see what's bothering you.
According to Munkres's definition, to say that $H\colon A\times [0,1]\to B$ is differentiable means that there is a neighborhood $U$ of $A\times [0,1]$ in $\mathbb R^{n+1}$ and a differentiable map $\widetilde H\colon U\to \mathbb R^m$ whose restiction to $A\times [0,1]$ is equal to $H$. The problem is that the image of $\widetilde H$ might not be contained in $B$, and therefore $\widetilde H^*\eta$ might not be defined on all of $U$.
But this is easily remedied. Because $B$ is open in $\mathbb R^m$ and $\widetilde H$ is continuous, the preimage $\widetilde H^{-1}(B)$ is an open subset of $\mathbb R^{n+1}$ containing $A\times [0,1]$. Let's call this preimage $U_0$, and let $H_0$ denote the restriction of $\widetilde H$ to $U_0$. Then $H_0$ is a differentiable map from a neighborhood of $A\times [0,1]$ into $B$, and $H_0^*\eta$ is a well-defined $(k+1)$-form defined on a neighborhood (namely $U_0$) of $A\times [0,1]$.
A: One point is that if $S$ is an arbitrary (in particular, not necessarily open) subset of $\mathbb R^m$ and $f: S\to \mathbb R^n$, then to say $f$ is differentiable on $S$ means that $f$ is the restriction of a differentiable function defined on an open neighborhood of $S$: Munkres spells this out in the definition on p199, just before Lemma 23.1. This neighborhood and extended function are often not mentioned explicitly, but in your situation the map $H$ is implicitly taken to be defined on an open neighborhood of $A\times [0,1]$. Hence $H^*\eta$ is also defined and smooth on this neighborhood, which is what it means to say $H^*\eta$ is smooth on the non-open set $A\times [0,1]$.
