$ \Phi : \operatorname{Hom}(U \otimes V,W) \to \operatorname{Hom}(U,\operatorname{Hom}(V,W)) $ Isomorphism Let $ U,V,W $ be $K$-vector-spaces.
Prove that $$ \Phi : \operatorname{Hom}(U \otimes V,W) \to \operatorname{Hom}(U,\operatorname{Hom}(V,W))  $$
$$  (\Phi(T)(u))(v) = T(u\otimes v) $$
with $ u \in U, v \in V $ is an isomorphism.
Can anyone help me with that?
 A: The OP had time to work the hints in the comments, so I will follow them to help people that just happens to pass over here :-)

(1) The "currying" $ \ Bilin(U \times V, W) \cong Hom \big( U, Hom(V,W) \big)$

Define $$\Delta: Bilin(U \times V, W) \to Hom \big( U, Hom(V,W) \big) \\ \nabla: Hom \big( U, Hom(V,W) \big) \to Bilin(U \times V, W) \\ \{[\Delta( \beta )] (u)\} (v) = \beta(u,v) \qquad \text{and} \qquad [\nabla(T)](u,v)= [T(u)](v)$$ for all $ \ u \in U \ $ and all $ \ v \in V$.
Check that in fact we have $ \ \Delta( \beta ) \in Hom \big( U, Hom(V,W) \big)$, $\forall \beta \in Bilin(U \times V, W)$, and $ \ \nabla(T) \in Bilin(U \times V, W)$, $\forall T \in Hom \big( U, Hom(V,W) \big)$. That is $ \ im(\Delta) \subset Hom \big( U, Hom(V,W) \big) \ $ and $ \ im(\nabla) \subset Bilin(U \times V, W)$. The definitions make sense.
So $ \ \{[(\Delta \circ \nabla) (T)](u)\}(v) = \big\{ \big[ \Delta \big( \nabla(T) \big) \big](u) \big\} (v) = [\nabla(T)](u,v)= [T(u)](v)$, $\forall v \in V$. Hence $ \ [(\Delta \circ \nabla) (T)](u) = T(u)$, $ \forall u \in U$. Then $ \ (\Delta \circ \nabla) (T) = T = id(T)$, $\forall T \in Hom \big( U, Hom(V,W) \big)$. Therefore $ \ \Delta \circ \nabla = id$.
For the other way around, $ \ [(\nabla \circ \Delta)(\beta)](u,v) = \big[ \nabla \big( \Delta (\beta) \big) \big] (u,v) = \{[\Delta (\beta)](u)\}(v) = \beta(u,v)$, $\forall (u,v) \in U \times V$. So $ \ (\nabla \circ \Delta)(\beta) = \beta = id(\beta)$, $\forall \beta \in Bilin(U \times V, W)$. Hence $ \ \nabla \circ \Delta = id$.
Then $ \ \nabla = \Delta^{-1} \ $ and both are bijective.
We have $ \ \{ [\Delta (\beta + \gamma)](u) \} (v) = (\beta + \gamma)(u,v) = \beta (u,v) + \gamma (u,v) = $ $ = \{ [\Delta (\beta)](u) \} (v) + \{ [\Delta (\gamma)](u)\} (v) = \{ [\Delta (\beta)](u) + [\Delta (\gamma)](u) \} (v)$ , $\forall v \in V$. So $ \ [\Delta (\beta + \gamma)](u) = [\Delta (\beta)](u) + [\Delta (\gamma)](u) = [\Delta (\beta) + \Delta (\gamma)](u)$, $\forall u \in U$. Hence $ \ \Delta (\beta + \gamma) = \Delta (\beta) + \Delta (\gamma)$, $\forall \beta, \gamma \in Bilin (U \times V, W)$.
Finally, $ \ \{ [\Delta (k \cdot \beta)](u) \} (v) = (k \cdot \beta)(u,v) = k \cdot \beta (u,v) = k \cdot \{ [\Delta (\beta)](u) \} (v) =$ $= \{ k \cdot [\Delta (\beta)](u) \} (v)$ , $\forall v \in V$. So $ \ [ \Delta (k \cdot \beta)](u) = k \cdot [\Delta (\beta)](u) = [k \cdot \Delta (\beta)](u)$, $\forall u \in U$. Hence $ \ \Delta (k \cdot \beta) = k \cdot \Delta (\beta)$, $\forall \beta \in Bilin (U \times V, W)$, $\forall k \in K$.
Then $\Delta$ is a homomorphism. Therefore $\nabla$ is a homomorphism too. So both are isomorphisms.

(2) The "universal" $ \ Bilin(U \times V, W) \cong Hom( U \otimes V ,W)$

For all $ \ \beta \in Bilin(U \times V, W)$, by definition of tensor products (universal property), there exists a unique $ \ \bar{\beta} \in Hom(U \otimes V, W) \ $ such that $ \ \bar{\beta} \circ \otimes = \beta$, where $ \ \otimes: U \times V \to U \otimes V \ $, $(u,v) \mapsto u \otimes v$, is the tensor (bilinear) map. Hence the relation $\mathsf{T}$ of pairs $$(\beta, \bar{\beta}) \in Bilin(U \times V, W) \times Hom( U \otimes V ,W)$$ given by this rule is a function $ \ \mathsf{T}: Bilin(U \times V, W) \to Hom( U \otimes V ,W) \ $ and $ \ \mathsf{T}(\beta) = \bar{\beta}$, $\forall \beta \in Bilin(U \times V, W)$. We have $ \ [(\bar{\beta} + \bar{\gamma}) \circ \otimes] (u,v) = (\bar{\beta} + \bar{\gamma}) (u \otimes v) =$ $= \bar{\beta} (u \otimes v) + \bar{\gamma} (u \otimes v) = (\bar{\beta} \circ \otimes) (u,v) + (\bar{\gamma} \circ \otimes) (u,v) = \beta (u,v) + \gamma (u,v) = (\beta + \gamma)(u,v)$, $\forall (u,v) \in U \times V$. So, $(\bar{\beta} + \bar{\gamma}) \circ \otimes = \beta + \gamma$. By uniqueness (in the universal property), $\mathsf{T}(\beta+\gamma) = \overline{\beta+\gamma} = \bar{\beta} + \bar{\gamma} = \mathsf{T}(\beta) + \mathsf{T}(\gamma)$, $\forall \beta , \gamma \in Bilin(U \times V, W)$. Now $ \ [(k \cdot \bar{\beta}) \circ \otimes] (u,v) = (k \cdot \bar{\beta}) (u \otimes v) = k \cdot \bar{\beta} (u \otimes v) = k \cdot (\bar{\beta} \circ \otimes) (u,v) =$ $= k \cdot \beta (u,v) = (k \cdot \beta) (u,v)$, $\forall (u,v) \in U \times V$. Then $ \ (k \cdot \bar{\beta}) \circ \otimes = k \cdot \beta$. Again by uniqueness, $\mathsf{T}(k \cdot \beta) = \overline{k \cdot \beta} = k \cdot \bar{\beta} = k \cdot \mathsf{T}(\beta)$, $\forall \beta \in Bilin(U \times V, W)$, $\forall k \in K$. Therefore $\mathsf{T}$ is a homomorphism.
Let $ \ \mathsf{G} : Hom( U \otimes V ,W) \to Bilin(U \times V,W) \ $ be such that $\mathsf{G}(f)= f \circ \otimes$, $\forall f \in Hom(U \otimes V, W)$. Use that $\otimes$ is bilinear and show that $\mathsf{G}(f) \in Bilin(U \times V,W)$, $\forall f \in Hom(U \otimes V, W)$. That is, $im(\mathsf{G}) \subset Bilin(U \times V,W) \ $ and the definition is right. For each $ \ f \in Hom( U \otimes V , W)$, by the universal property, there is a unique $ \ \overline{\mathsf{G}(f)} = \mathsf{T} \big( \mathsf{G}(f) \big) \in Hom(U \otimes V, W) \ $ such that $ \ \overline{\mathsf{G}(f)} \circ \otimes = \mathsf{G}(f)$. But $ \ f \in Hom( U \otimes V , W) \ $ and $ \ f \circ \otimes= \mathsf{G}(f)$. So, by uniqueness, $(\mathsf{T} \circ \mathsf{G})(f) = \mathsf{T} \big( \mathsf{G}(f) \big) = \overline{\mathsf{G}(f)} = f = id(f)$. Hence $ \ \mathsf{T} \circ \mathsf{G} = id$. For the other way around, again by the universal property, we have $ \ (\mathsf{G} \circ \mathsf{T})(\beta) = \mathsf{G} \big( \mathsf{T} (\beta) \big) = \mathsf{G} ( \bar{\beta}) = \bar{\beta} \circ \otimes = \beta = id(\beta)$, $\forall \beta \in Bilin(U \times V, W)$. Hence $ \ \mathsf{G} \circ \mathsf{T} = id \ $ and both $\mathsf{T}$ and $\mathsf{G}$ are bijective, with $ \ \mathsf{G} = \mathsf{T}^{-1}$. Therefore $\mathsf{G}$ is a homomorphism and both $\mathsf{T}$ and $\mathsf{G}$ are isomorphisms.
Obs.: Let $ \ f \in Hom(U \otimes V,W)$. We have that $ \ \{ [(\Delta \circ \mathsf{G})(f)](u) \} (v) = $ $= \big\{ \big[ \Delta \big( \mathsf{G}(f) \big) \big] (u) \big\} (v) = \{ [ \Delta(f \circ \otimes)] (u) \} (v) = (f \circ \otimes) (u,v) = f(u \otimes v) = \{ [\Phi(f)](u) \} (v)$, $\forall v \in V$. So, $[(\Delta \circ \mathsf{G})(f)](u) = [\Phi(f)](u)$, $\forall u \in U$. Hence $ \ (\Delta \circ \mathsf{G})(f) = \Phi(f)$. Since $f$ is arbitrary, the OP function is $ \ \Phi = \Delta \circ \mathsf{G}$, a isomorphism.
A: Let $f: M \times N \to P$ be a bilinear map. Then for every $x \in M$, the map $y \mapsto f(x,y):N \to P$ is also linear. This gives a map $M \to \mathrm{Hom}(N,P)$. Conversely, a map $\phi:M \to \mathrm{Hom}(N,P)$ defines a bilinear map $(x,y) \mapsto \phi(x)(y): M \times N \to P$. This means that bilinear maps $\{f:M \times N \to P\} \iff \{\mathrm{Hom}(M,\mathrm{Hom}(N,P)\}$.
Of course we also have that $\mathrm{Hom}(M \otimes N,P) \iff \{f:M \times N \to P\}$.
Thus, the isomorphism is given by taking  $ \phi \in\mathrm{Hom}(M \otimes N,P)$ and sending it to $y \mapsto \phi(x \otimes y)$, which is a bijection by the argument above.
