Extension of scalars of the dual vector space I was working through the following https://math.stackexchange.com/a/426300/299525 and I could not justify one of the steps. 
It seems to be a consequence of a general result, as discussed in the comments.

Proposition : If $K/L$ is an extension of fields, $V$ a vector space over $L$, then
  $$ \textrm{Hom}_L (V,K) \cong \textrm{Hom}_L (V,L) \otimes _L K$$

So it seems most natural to define a map going the other way out of the tensor product, perhaps $(\varphi , k) \mapsto k (\psi \circ \varphi)$, where $\psi : L \to K$. This is well-defined and bilinear so it induces a map from the tensor. Injectivity is also clear since $\psi$ is an injection. 
However, I am not sure why every $L$-linear homomorphism $V \to K$ must factor through $\psi$, up to a multiple of $K$ though, i.e. why it should be surjective. 
 A: This is only true (at least in a natural way) if either $K$ or $V$ is finite-dimensional over $L$.  If $K$ is finite-dimensional, pick a basis $\{e_1,\dots,e_n\}$ of $K$.  Given a linear map $\alpha:V\to K$, write $\alpha(v)=\alpha_1(v)e_1+\dots+\alpha_n(v)e_n$ with $\alpha_i(v)\in L$.  Then each $\alpha_i$ is a linear map $V\to L$, and $\alpha$ is the image of $\alpha_1\otimes e_1+\dots+\alpha_n\otimes e_n$ under the map you have described.
If $V$ is finite-dimensional, pick a basis $\{v_1,\dots,v_n\}$ for $V$, and let $\{\alpha_1,\dots,\alpha_n\}$ be the dual basis of $\operatorname{Hom}_L(V,L)$, so $\alpha_i(v_j)=\delta_{ij}$.  Then any $\alpha:V\to K$ is the image of $\alpha_1\otimes \alpha(v_1)+\dots+\alpha_n\otimes\alpha(v_n)$ under the map you have described.
(Note that in particular, it is not true that every map $V\to K$ factors through $\psi$, up to a multiple of $K$.  Rather, every map is a linear combination of maps that factor through $\psi$.  This is not true when $K$ and $V$ are both infinite-dimensional over $L$, essentially because you can't take "infinite linear combinations".)
