Is the Grassmann Functor $\Lambda$ full?

Question

Is the functor $\Lambda: \mathsf{FinDimVect}_\mathbb{R} \to \mathsf{Alg}_\mathbb{R}$ that sends a fin. dim. $\mathbb{R}$-vector space to its exterior algebra full? If not, is there a way of constructing an arbitrary Grassmann algebra homomorphism systematially (maybe as a linear combination of functor images of linear maps)?

Answer 1

Given a $\Bbb Z/2\Bbb Z$-graded-commutative $\Bbb R$-algebra $A=A^0\oplus A^1$, that is $A^iA^j\subset A^{i+j}$ and for all $a\in A^i$, $b\in A^j$, $ba=(1)^{ij}ab$, there is a one to one correspondence between algebra homomorphisms $\Lambda V\to A$ and linear maps $V\to A^1$.

In particular, for $A=\Lambda W=\Lambda^{\text{even}} W\oplus\Lambda^{\text{odd}}W$, algebra homomorphisms $\Lambda V\to \Lambda W$ are in one to one correspondence with linear maps $V\to\Lambda^{\text{odd}}W$. The ones you get from the "exterior algebra functor" are only those in the subclass of algebra homomorphisms that arise from a map $V\to W\big(\hookrightarrow\Lambda^{\text{odd}}W\big)$.