Weird condition implies isomorphism between vector spaces?

Question

Let $U$ and $V$ be vector spaces. It can be shown that the existence of atleast one space $W$ such that $\mathscr L(U,W) \cong \mathscr L(V,W)$ is not sufficient for $U$ and $V$ to be isomorphic as, for example, if we consider $\mathbb R[X]$ as the space of all real polynomials, it is true that $End(\mathbb R[x]) \cong \mathscr L(\mathbb R,\mathbb R[X])$ although it is obviously not isomorphic to $\mathbb R$. My question is: If we strengthen the condition and require that for ANY vector space $W$ we have that $\mathscr L(U,W) \cong \mathscr L(V,W)$, does that imply that $U \cong V$? This is a way more powerful restriction and so I can imagine the proposition would hold, specially since it seems almost impossible to think of any possible counter-examples to this. What seems like a plausible thing to do is to try to extract some argument about the cardinality of the basis of $U$ and $V$, but it's not entirely clear to me how to do that.

Answer 1

this is an example of "Yoneda lemma". it is a general principle wich says that $HOM(X,-)$(or $Hom(-,X)$) is always an injection from any category to the category of sets.so if there are functorial isomorphism between $Hom(U,W)=Hom(V,W)$ for each $W$ then $U,V$ are isomorphism.(functorial means that the isomorphism are compatible with maps $W\to W'$)

for the proof just take $W=V,U$ you get a map $U\to V$ and a map in the inverse direction(by taking inverse image of identity) and functoriality says that this morphism are inverse to each other