Wedge product of maps If $V$ and $W$ are $\mathbb{F}$ vector spaces, A $k$-multilinear alternating map $V^k \to W$ induces a unique linear map $f: \bigwedge^k V \to W$.

In the special case $W = \mathbb{F}$ and $\dim V &lt; \infty$ (correct me if I'm wrong, I believe this last definition is necessary), we can write down $f$ as a sum of elements like $f_1 \wedge \dotsb \wedge f_k$, where $f \in V^*$. When $W \neq \mathbb{F}$, is there any notion of the wedge product of maps $\phi: V \to W$? How would it be evaluated?

Or is the best we can do to write
$$
f = \sum_{i=1}^n ((f_1^i \wedge \dotsb \wedge f_k^i)\otimes w_i), \quad \text{for } f_j^l \in V^*?
$$
Let $\{w_i\}_{i\in I}$ be a basis for $W$. 

Let $\{w_i^*\in W^*\}_{i\in I}$ be the dual set, i.e., $w_i^*:W\rightarrow\mathbb{F}$ is  a linear functional s.t. $w_i^*(w_j)=\delta_{ij}$. 

Since every $w\in W$ can be written as a finite linear combination of the elements of this basis then $w=\sum_{i\in I}w_i^*(w)w_i$ (Notice that this is a finite sum).

Thus, you can write the linearization $F:\bigwedge^kV\rightarrow W$ as $F(r)=\sum_{i\in I}w_i^*(F(r))w_i$. 

Now, the map $w_i^*\circ F: \bigwedge^kV\rightarrow \mathbb{F}$  can be described as you know.
Let $\{w_i\}_{i\in I}$ be a basis for $W$. 

Let $\{w_i^*\in W^*\}_{i\in I}$ be the dual set, i.e., $w_i^*:W\rightarrow\mathbb{F}$ is  a linear functional s.t. $w_i^*(w_j)=\delta_{ij}$. 

Since every $w\in W$ can be written as a finite linear combination of the elements of this basis then $w=\sum_{i\in I}w_i^*(w)w_i$ (Notice that this is a finite sum).

Thus, you can write the linearization $F:\bigwedge^kV\rightarrow W$ as $F(r)=\sum_{i\in I}w_i^*(F(r))w_i$. 

Now, the map $w_i^*\circ F: \bigwedge^kV\rightarrow \mathbb{F}$  can be described as you know.