Let $U$ and $V$ be $\mathbb{F}$-vector spaces, for some field $\mathbb{F}$. I want to prove that $U \otimes_{\mathbb{F}} V \cong \text{Hom}_{\mathbb{F}}(U', V)$. Here, I am using tensor product of vector spaces, $\text{Hom}_{\mathbb{F}}(U', V) := \{T : U' \to V \, \mid \, T \text{ is linear with scalars in } \mathbb{F}\}$, and $U'$ is the dual space of $U$.
Let $\varphi: U \otimes_{\mathbb{F}} V \to \text{Hom}_{\mathbb{F}}(U', V)$ such that $\varphi(u \otimes v) = M$ with $$M: U' \to V \text{ such that } m \mapsto (m,u) \cdot v$$.
However, using these notations, how can I show linearity, injectivity, and surjectivity?
I am a beginner in tensor products, so please write simple answers.
Thanks.
First, this is not true in general. Take $\mathbb F = \mathbb F_2$, $U=\oplus_{i\in\mathbb N}\mathbb F_2$ and $V=\mathbb F_2$. Then clearly $U\otimes_{\mathbb F} V$ has countable dimension, but $\text{Hom}_{\mathbb F_2}(U', V)$ has uncountable dimension: $U'=\text{Hom}_{\mathbb F_2}(U, \mathbb F_2)$ already has uncountable dimension.
In the finite dimensional case, first show $\tilde{\phi}:U\times V\rightarrow \text{Hom}(U', V)$ defined by $\phi(u, v)(f) = f(u)v$ is bilinear, hence induces a homomorphism $\phi:U\otimes_{\mathbb F} V\rightarrow \text{Hom}(U', V)$. Now we can show $\phi$ is surjective: Pick basis $\{e_i\}$ of $U$ and dual basis $\{e_i^*\}$ of $U'$. For any $f\in\text{Hom}(U', V)$, we can check $\sum_i e_i\otimes f(e_i^*)$ is mapped to $f$ since $\phi(\sum_i e_i\otimes f(e_i^*))(e_j^*)=\sum_ie_i(e_j^*)f(e_i^*)=f(e_j^*)$, that is $\phi(\sum_i e_i\otimes f(e_i^*))$ and $f$ agree on a basis, hence must be the same.
Now since $U\otimes V$ and $\text{Hom}(U', V)$ have the same dimension, $\phi$ must be an isomorphism.