Let $V$ and $W$ be finite dimensional $K$ vector spaces. Prove that for $\varphi \in V^*$ and $\psi \in W^*$ exists a well-defined map: $$P_{\varphi , \psi} : V \otimes W \to K, v\otimes w \mapsto \varphi(v) \psi(w)$$
I would like some help in understanding what is meant by this assignment in one of the Linear Algebra II books that i own. I have been trying to figure out how we can find a map like this. I already have a few ideas:
We can use $\varphi$ and $\psi$ on $v$ and $v$ before using the Tensor Product, then try to use the multilinear property to obtain the map that was given above. Though i got the hunch, that i did not understand the properties of Tensor Products with maps/functional's at all.
Any help is appreciated.
The $K$-linear space $V \otimes W$, together with the bilinear map $V \times W \xrightarrow{\otimes} V \otimes W$ such that $(v,w)\mapsto v \otimes w$, is characterised (up to isomorphism of $K$-linear spaces) by the following universal property (the one that you are talking about in your question): whenever $U$ is a $K$-linear space and $f \colon V \times W \to U$ is a bilinear map, there is unique a $K$-linear map $f'\colon V\otimes W \to U$ such that $f' \circ \otimes = f$.
Hence in this case, in order to get that your map $P_{\phi,\psi}$ (which will be your $f'$) exists and is $K$-linear, you only need to observe that the corresponding arrow: $$f\colon V \times W \ni (v,w)\mapsto \phi(v)\psi(w) \in K$$ is bilinear (here you'd use that $\phi$ and $\psi$ are elements of the dual spaces, that is, they are $K$-linear). Then you're done by the universal property of $V\otimes W$ that we are talking about.