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)?
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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 $V\to W\big(\hookrightarrow\Lambda^{\text{odd}}W\big)$.
Olivier Bégassat
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Thank you very much! so the above holds just for the case $V=W=\mathbb{R}$ – user154917 Jun 03 '14 at 13:30
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It will work (for all $V$) when $W$ two dimensional. – Olivier Bégassat Jun 03 '14 at 13:32
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ah, yes of course – user154917 Jun 03 '14 at 13:35
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the algebra homomorphisms in this correspondence preserve the homogeneous degree, right? – user154917 Jun 03 '14 at 14:05
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Certainly not! Those that do are precisely those that arise from maps $V\to W$. They do however conserve the parity of the degrees. – Olivier Bégassat Jun 03 '14 at 14:18
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Sorry, with homogeneous degree i meant the degree in the $\mathbb{Z}_2$ gradation.. – user154917 Jun 03 '14 at 14:22
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So why not work with $\mathbb{Z}$-graded algebras, then? – Qiaochu Yuan Jun 03 '14 at 18:03