Finding unique linear transformation in tensor product.

I'm trying to learn tensor products by myself, and I'm using primarily the Linear Algebra and Geometry book, by Kostrikin and Manin. I need help with the following question:

Let $$V$$ and $$W$$ be $$\mathbb{K}$$-vector spaces. Show that there is a unique linear transformation $$\Gamma: V \otimes W \rightarrow \mathcal{L}(V^*,W)$$ that satisfies $$\Gamma(v\otimes w)(f) = f(v)w$$ for any $$v \in V, w \in W$$ and $$f\in V^*$$.

My first (and only) idea was to define an application $$t$$ from $$V\times W$$ to $$V \otimes W$$ that associates the pair $$(v,w)$$ to $$v\otimes w$$, and use (somehow) that $$t$$ is universal. However, I'm very much stuck.

Any help would be deeply appreciated.

• Assuming $\mathcal{L}(V^*,W)$ means linear maps $V^*\to W$ then define $\Gamma:V\times W\rightarrowtail \mathcal{L}(V^*,W)$ by $\Gamma(v,w):f\mapsto f(v)w$. Prove $\Gamma$ is biadditive (distributes) and apply universal mapping property of the tensor product. Jul 25 '19 at 19:50
• Thank you, @Algeboy. I've managed to answer the question. If no one post an answer, I will post mine latter on, just to close the question. Jul 25 '19 at 20:17

Recall that every pair of $$\Bbb K$$-vector spaces $$V$$ and $$W$$ defines a map where $$\Phi$$ is the $$\Bbb K$$-bilinear map defined by $$\Phi(v, w)=v\otimes w$$. The map $$\Phi$$ is characterized by the following universal property.

Universal Property of $$\Phi$$. Consider a diagram of $$\Bbb K$$-vector spaces where $$\phi$$ is $$\Bbb K$$-bilinear. Then there exists a unique $$\Bbb K$$-linear map $$\Gamma$$ making commute.

Next, to address your example, take $$X=\mathcal L(V^\ast, W)$$. The space $$V^\ast$$ consists of all linear maps $$V\to \Bbb K$$ and the space $$\mathcal L(V^\ast, W)$$ consists of all linear maps $$V^\ast \to W$$.

Now, let $$\phi:V\times W\to\mathcal L(V^\ast, W)$$ be defined by $$\phi(v, w)(f)=f(v)\cdot w$$. The formulas \begin{align*} \phi&(\lambda_1\cdot v_1+\lambda_2\cdot v_2, w)(f)\\ &= f(\lambda_1\cdot v_1+\lambda_2\cdot v_2)\cdot w & \phi&(v, \lambda_1\cdot w_1+\lambda_2\cdot w_2)(f)\\ &= f(v)\cdot(\lambda_1\cdot w_1+\lambda_2\cdot w_2) \\ &= \{\lambda_1\cdot f(v_1)+\lambda_2\cdot f(v_2)\}\cdot w & &= \lambda_1\cdot f(v)\cdot w_1+\lambda_2\cdot f(v)\cdot w_2 \\ &= \lambda_1\cdot f(v_1)\cdot w+\lambda_2\cdot f(v_2)\cdot w & &= \lambda_1\cdot\phi(v, w_1)(f)+\lambda_2\cdot\phi(v, w_2)(f) \\ &= \lambda_1\cdot\phi(v_1, w)(f)+\lambda_2\cdot\phi(v_2, w)(f) \end{align*} demonstrate that $$\phi$$ is $$\Bbb K$$-bilinear.

Now, we have a diagram where $$\phi$$ is bilinar. What does the universal property of $$\Phi$$ say?