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Let $k$ be a field, let $V$ and $W$ be $k$-vector spaces of dimensions $n$ and $m$ respectively, and let $f:V\to W$ be a $k$-linear transformation. Let $\Lambda(V)$ and $\Lambda(W)$ denote the exterior algebras of $V$ and $W$ respectively. So we have $\Lambda(V) = \Lambda^0(V)\oplus\Lambda^1(V)\oplus\cdots\oplus\Lambda^n(V)$ and $\Lambda(W) = \Lambda^0(W)\oplus\Lambda^1(W)\oplus\cdots\oplus\Lambda^m(W).$

The wikipedia page on exterior algebras states that there is a unique function $\Lambda(f):\Lambda(V)\to\Lambda(W)$ such that $\Lambda(f)|_{\Lambda^1(V)}:\Lambda^1(V)\to\Lambda^1(W)$ is defined by $\Lambda(f)(v)=f(v)$.

In fact, $\Lambda(f)$ preserves grading (i.e. it can be written as a sum of maps $\Lambda^k(f):=\Lambda(f)|_{\Lambda^k(V)}:\Lambda^k(V)\to\Lambda^k(W)$). If $1\leq k \leq n$, then $\Lambda(f)$ is given by $\Lambda^k(f)(v_1\wedge\cdots\wedge v_k) = f(v_1)\wedge\cdots\wedge f(v_k).$

I do not understand how this function acts on $\Lambda^0(V)=k$. I know that we have a map $\Lambda^0(f):\Lambda^0(V)\to\Lambda^0(W)$ which is really the same as $\Lambda^0(f):k\to k.$

My question has two parts: what is $\Lambda^0(f)$ and how is it determined from the universal mapping property for exterior algebras?

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    @Arturo: Thanks. I promptly upvoted (and I didn't care about the CW), and I see that a couple of others have since joined in:) (reposted to fix a typo)2011-04-02

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The map is the identity map. The map must be $k$-linear, so $\Lambda^0(f)(a) = \Lambda^0(f)(a\cdot 1) = a\Lambda^0(f)(1)$ hence the map is completely determined by the image of $1$. But since it is a morphism of algebras, the map must send $1$ to $1$, so $\Lambda^0(f)(1)=1$, hence $\Lambda^0(f)(a) = a$ for all $a\in k$.