Differential forms on standard simplices via Whitney extension vs diffeological structure The standard simplices $\Delta^n \subset \{\mathbf{x}\in\mathbb{R}^{n+1}\mid x_0 + \ldots + x_n  =1 \} =: \mathbb{A}^n$ carry two natural sorts of smooth differential forms:


*

*Those differential forms on the interior of $\Delta^n$ that extend smoothly to a neighbourhood of $\Delta^n$ in $\mathbb{A}^n$ (this definition is used for instance by Dupont in his book Curvature and Characteristic Classes). This is implicitly using Whitney's extension theorem, I think.

*Consider $\Delta^n$ as a diffeological subspace of $\mathbb{A}^n$, where a function $\phi\colon \mathbb{R}^k \to \Delta^n$ is a plot if and only if the composite $\mathbb{R}^k \to \Delta^n \hookrightarrow \mathbb{A}^n$ is smooth. A differential form $\omega$ on $\Delta^n$ then consists of the data of a differential form $\phi^*\omega$ for each plot $\phi$ with a compatibility condition when a plot factors through another via a smooth map between Euclidean spaces.
The first of these is more of a "maps out" viewpoint, and probably corresponds to a natural smooth space structure defined via smooth real-valued functions. The second is a "maps in" viewpoint. Note that the $D$-topology arising from the diffeology in 2. above is the standard topology on the simplex. We then get a cochain complex of differential forms of each type, as exterior differentiation can be defined in the more-or-less obvious way in each case.

My question is: how do these relate? Is one a subcomplex of the other? Or are they quasi-isomorphic, via a third cochain complex?

The motivation is that Dupont's simplicial differential forms on semisimplicial manifolds $X_\bullet$ look like they should be differential forms on the fat geometric realisation considered as a diffeological space, since a simplicial differential form is more or less descent data for the sheaf of differential forms and the 'cover' $\coprod_{n\geq 0} \Delta^n\times X_n \to ||X_\bullet||$, assuming the first definition above. However, if his differential forms on $\Delta^n$ (or more precisely on $\Delta^n\times X_n$) aren't diffeological differential forms, they don't give a form on $||X_\bullet||$ as a diffeological space. I guess all we really need is a map of cochain complexes from the first to the second given above.
 A: The two chain complexes are isomorphic for any $n≥0$.
Fix some $n≥0$, $k≥0$ and consider $k$-forms on the $n$-simplex.
I will use the notations $Ω_e^k$ and $Ω_d^k$ for forms of type 1 and 2 respectively.
I also use the notation $Δ_d^n$ for the diffeological $n$-simplex
so that $\def\Hom{\mathop{\rm Hom}} \def\R{{\bf R}} \def\d{\,{\rm d}} Ω_d^k=\Hom(Δ_d^n,Ω^k)$.
First, we have a canonical map $ι\colon Ω_e^k→Ω_d^k$ that
sends $ω∈Ω_e^n$ to the map $Δ_d^n→Ω^k$
that sends a smooth map $φ\colon S→Δ_d^n$ to the $k$-form $φ^*ω∈Ω^k(S)$.
Secondly, the map $ι$ is injective,
which follows from the following two observations.
By the Yoneda lemma the restriction of $ι(ω)\colon Δ_d^n→Ω^k$ to the sheaf represented by the open interior of $Δ_d^n$ equals the restriction of $ω$ to the open interior of $Δ^n$.
Furthermore, any two forms in $Ω_e^k$ whose restrictions to the open interior of $Δ^n$ coincide must be equal (by continuity).
Thirdly, the map $ι$ is surjective.
This can be shown using a two-step construction:
in the first step we construct a certain element $ω$ of $Ω_e^k$
starting from some given element $ψ$ of $Ω_d^k$
and in the second step we show that $ι(ω)=ψ$.
First, let's briefly examine the simplest nontrivial case $d=n=1$.
We have a form $ψ∈Ω_d^1(Δ_d^1)=Ω_d^1([0,1])$,
and we already know its restriction $f(x) \d x$ to the open interval $(0,1)$.
In particular, $f$ is a smooth function on $(0,1)$.
Pulling back $ψ$ along the plot $\R→\R$ ($x↦x^2$)
yields some 1-form $g(x) \d x$, where for any $x≠0$ we have $g(x)=2xf(x^2)$
and $g$ is a smooth function on $(-1,1)$.
Since $g$ is odd, we have $g(0)=0$,
so the even function $h$ given by $h(x)=g(x)/(2x)$ is smooth on $(-1,1)$
and we have $h(x)=f(x^2)$ for all $x≠0$.
Set $f(0)=h(0)$.
We claim $f$ is smooth on $(-1,1)$.
Indeed, $h'(x)=2xf'(x^2)$ for $x≠0$,
and since $h'$ is odd, we have $h'(0)=0$
and $h'(x)/(2x)$ is a smooth function on $(-1,1)$
whose value at $x=0$ is $f'(0)$.
We repeat this step by induction, proving $f$ has derivatives of all orders at 0.
The general case is nothing else than a multivariable paremetrized
version of the above argument.
More precisely, we argue as follows.
For the first step, suppose we are given some $ψ∈Ω_d^k$.
Given some point $x∈Δ^n$, we would like to define $ω(x)$.
Denote by $d≥0$ the codimension of the stratum of $Δ^n$ that contains $x$.
Parametrize the simplex $Δ^n$ using $n$ coordinates
in such a way that $x_1=⋯=x_d=0$
and the other $n-d$ coordinates of $x$ are strictly positive,
so that the points $y$ of the stratum containing $x$ are defined by the relations $y_i≥0$ for $1≤i≤d$.
Decompose $$ω=∑_I g_I ∏_{i∈I} \d x_i$$ into its individual components with respect to this coordinate system.
Pick one such component $g_I ∏_{i∈I} \d x_i$; we would like to define $g_I$ as a smooth function on some open neighborhood $U$ of the given stratum of $Δ^n$, which is parametrized by the remaining $n-d$ coordinates $x_{d+1}$, …, $x_n$.
Consider the smooth map $U→Δ^n$ that (in the coordinates introduced above)
sends $x_i↦x_i^2$ for $1≤i≤d$ and $x_i↦x_i$ for $i>d$.
This defines a morphism $τ\colon U→Δ_d^n$, so we have a form $τ^*ψ∈Ω^k(U)$.
Now take the coefficient $h$ of $τ^*ψ$ before $∏_{i∈I} dx_i$.
Take the partial derivative of $h$ with respect to all coordinates $x_i$
such that $i∈I$ and $i≤d$.
Divide the resulting function by $2^{\#(I∩\{1,…,d\})}$.
This is the function $g_I$.
This argument also proves that the resulting $k$-form $ω$ is smooth.
For the second step, we have to show that $ι(ω)=ψ$.
Reusing the notation of the previous paragraph,
consider some arbitrary plot $φ\colon S→Δ_d^n$ such that $φ(s)=x$ for some $s∈S$.
The first $d$ coordinates of $φ$ must be nonnegative in a neighborhood $U$ of $s$.
Since $φ$ is differentiable, the first derivatives of these $d$ coordinates must vanish at the point $s$.
Thus, taking the square root of each of the first $d$ coordinates produces a smooth map $ε\colon U→\R^n$ such that $φ=τε$, where the map $τ$ was defined in the previous paragraph.
Now $φ^*ι(ω) = ε^* τ^* ι(ω) = ε^* τ^*ψ = φ^*ψ$,
where $τ^*ι(ω)=τ^*ψ$ by definition of $ω$.
