Is the de Rham complex a free (commutative?) differential graded algebra? A differential graded algebra (dg-algebra) is a monoid object in the category of chain complexes with respect to the usual tensor product of complexes. A (graded) commutative dg-algebra is simply a commutative monoid object.
The de Rham complex for a smooth manifold is easily seen to be a commutative dg-algebra with respect to the wedge product.
I could have sworn that I saw somewhere that the de Rham complex is the free dg-algebra (or free commutative dg-algebra?) on some appropriate base category, i.e. left adjoint to a forgetful functor. I can't seem to find a reference for this, however. 
With regards to my background I'm familiar with category theory and differential geometry, but my algebra is a bit weak.
 A: I think it is true that the algebraic de Rham complex is left adjoint to the forgetful functor from DGAs to $R$-algebras. 

More precisely:
$\newcommand{\dgAlg}{\mathrm{dgAlg}}
\newcommand{\Alg}{\mathrm{Alg}}\newcommand{\Hom}{\mathrm{Hom}}$
Proposition 1. We have an adjunction

where


*

*$\Omega^\bullet_{(-)/A}$ is the exterior algebra of the module of Kähler differentials of an $A$-algebra $A\longrightarrow(-)$, and

*$忘\colon\dgAlg_A\longrightarrow\Alg_A$ is the forgetful functor sending a DGA $(D^\bullet,d^\bullet,\mu^{\bullet,\bullet})$ to the $A$-algebra $(D^0,\mu^{0,0})$.
In other words, the algebraic de Rham complex is the free differential graded algebra.

Now, for the proof.
Lemma 1 (Universal Property of the $p$th Exterior Power). Let $R$ be a ring and $M$ be an $R$-module. The pair $(\bigwedge^{p}M,p)$ where $p\colon M^{\otimes p}\twoheadrightarrow\bigwedge^{p}M$ is the natural projection satisfies the following universal property: 


*

*Given another pair $(N,f)$ where $N$ is an $R$-module and $f\colon M^{\otimes k}\longrightarrow N$ is an alternating $R$-linear morphism of $R$-modules, there exists a unique morphism $\bigwedge^{p}M\dashrightarrow N$ of $R$-modules such that the diagram



commutes.
Proof of Proposition 1. We proceed by showing that we have a functorial isomorphism
$$\Hom_{\dgAlg_A}(\Omega^\bullet_{B/A},D^\bullet)\cong\Hom_{\Alg_A}(B,D^0)$$
for each $A$_algebra $B$ and each differential graded algebra $(D^\bullet,d^\bullet,\mu^{\bullet,\bullet})$.
Firstly, given a morphism $f^\bullet\colon\Omega^\bullet_{B/A}\longrightarrow D^\bullet$ of differential graded algebras, we get a morphism $f^0\colon B\longrightarrow D^0$ of $A$-algebras given as the $0$th component of $f^\bullet$.
Conversely, let $f\colon B\longrightarrow D^0$ be a morphism of $A$-algebras and consider the diagram

We claim that there exists a unique morphism $f^\bullet\colon\Omega^\bullet_{B/A}\longrightarrow D^\bullet$ of differential graded algebras such that $f^0=f$. This we prove via universal properties:
The Degree $1$ Case. Consider the diagram
,
note that $d\circ f$ is a $B$-derivation into $D^1$, and apply the universal property of the module of Kähler differentials (see here) to get a unique map $\Omega_{B/A}^1\dashrightarrow D^1$ making the diagram

commute. This gives the degree $1$ component of our desired map $f^\bullet$ and shows the first square of the diagram given in the second picture of this answer to be commutative.
The Degree $p$ Case. Write $\Omega_{B/A}^1\otimes_A\cdots\otimes_A\Omega_{B/A}^1$ for the $p$th tensor product of $\Omega_{B/A}^1$ and consider the morphism
$$f^1\otimes_A\cdots\otimes_Af^1\colon\Omega_{B/A}^1\otimes_A\cdots\otimes_A\Omega_{B/A}^1\longrightarrow D^1\otimes_A\cdots\otimes_AD^1.$$
By functoriality of the $p$th exterior power, there is a unique morphism from $\Omega_{B/A}^p$ to $\bigwedge^pD^1$ such that the diagram

commutes. The multiplication maps $\mu^{p,q}\colon D^p\times D^q\longrightarrow D^{p+q}$ assemble into a unique multilinear map
$$(D^1)^p\longrightarrow D^p,$$
(the uniqueness of $\mu$ follows from the compatibility of the $\mu^{p,q}$'s) which corresponds, by the universal property of tensor products (i.e. that $R$-multilinear maps from $M^p$ correspond to $R$-linear maps from $M^{\otimes p}$), to a unique linear map
$$\mu\colon D^1\otimes_A\cdots\otimes_A D^1\longrightarrow D^p.$$
Moreover, the graded Leibniz rule for the $\mu^{n,m}$'s implies that $\mu$ is alternating. Consider now the diagram

By the universal property of the $p$th exterior power (Lemma 1), there exists a unique map $\bigwedge^{p}D^1\dashrightarrow D^{p}$ making the diagram

commute. Our desired $p$th component $f^p$ of $f^\bullet$ is then the composition of the vertical rightmost morphisms in this last diagram.
Thus we get a unique morphism $f^\bullet$ such that $f^0=f$. Since the constructions given here were done via universal properties, it is clear that they are inverse and functorial. This finishes the proof.
