Two defnitions of a tensor There are two approaches to define a tensor:  1) As an element of tensor-product space
2) As a multilinear map
So basically there is some correspondence between those definitions. 
However, I can't see how it can be done in a proper way.
Question: For example , given multilinear map $T:V^*\times V^*\to \mathbb{R}$ , then how do we construct correspondent $t\in V\otimes V$?
And other way around: given element $t'\in V\otimes V$, how do we construct correspondent $T':V^*\times V^*\to \mathbb{R}$?
My attempt: Let $T:V^*\times V^*\to \mathbb{R}$ be a given map $(dimV<\infty)$.
Using this map we can construct isomorphism $\phi_T:V^*\to V$ as: $$(\omega)\mapsto\phi_T(\omega):=T(\omega,\cdot)$$ 
Then we can construct isomorphism: $\Phi_T:V^*\times V^*\to V\times V$ as:
$$(\omega,\sigma)\mapsto \Phi_T(\omega,\sigma):=(\phi_T(\omega),\phi_T(\sigma))$$
Thus we have a map: $f:V\times V\to\mathbb{R}$ such that $T=f\circ\Phi_T$, or other way $f=T\circ\Phi^{-1}_T$. 
Finally that we have $V\times V$ and map $f$ we can use universal property and construct tensor space $V\otimes V$ with two maps: quotient map $g:V\times V\to V\otimes V$, and map $h:V\otimes V\to \mathbb{R}$, such that $f=h\circ g$ holds.
Given all of this one can see that: $T=h\circ g\circ \Phi_T$.
So for any $\omega,\sigma \in V^*$:
$$T(\omega,\sigma)=(h\circ g)(\phi_T(\omega),\phi_T(\sigma))=h(g(\phi_T(\omega),\phi_T(\sigma)))=h(g(T(\omega,\cdot),T(\sigma,\cdot)))$$
Then we can denote $g(T(\omega,\cdot),T(\sigma,\cdot))\equiv T(\omega,\cdot)\otimes T(\sigma,\cdot)$ and call it tensor generated by map $T$. 
Is this construction viable? Also, I'm not really sure how to go other way around and construct bilinear map from the element of tensor product.
 A: If $V_1, \ldots , V_k$ are finite dimensional vector spaces, then consider the function $$\phi : V_1^* \times \ldots \times V_k^* \rightarrow L(V_1, \ldots , V_k;\mathbb{R}),$$ $L(V_1, \ldots , V_k;\mathbb{R})$ the space of multilinear functions from $V_1 \times \ldots \times V_k$  to $\mathbb{R}$, such that $$(w_1,\ldots,w_k) \mapsto\phi(w_1,\ldots,w_k)$$ $$\phi(w_1,\ldots,w_k)(v_1,\ldots,v_k) = w_1(v_1)\ldots w_k(v_k)$$ Note that $\phi(w_1,\ldots,w_k) \in L(V_1, \ldots , V_k;\mathbb{R})$ and that $\phi$ is multilinear. Using the Universal Property for Tensor Product Spaces there exists a unique function $$\tilde{\phi}:V_1^* \otimes \ldots \otimes V_k^* \rightarrow L(V_1, \ldots , V_k;\mathbb{R})$$ such that $\tilde{\phi} \circ \pi = \phi$ with $\tilde{\phi}$ linear and $\pi: V_1^* \times \ldots \times V_k^* \rightarrow V_1^* \otimes \ldots \otimes V_k^*$ the projection taking each $(w_1,\ldots,w_k)$ to $w_1 \otimes \ldots \otimes w_k$. Note that $$\tilde{\phi}(w_1 \otimes \ldots \otimes w_k) = \tilde{\phi}(\pi (w_1, \ldots, w_k)) = \phi(w_1,\ldots,w_k) \in L(V_1, \ldots , V_k;\mathbb{R})$$ (notation $w_1 \otimes \ldots \otimes w_k := \phi(w_1,\ldots,w_k)$), thus $$\tilde{\phi}(w_1 \otimes \ldots \otimes w_k) = w_1 \otimes \ldots \otimes w_k.$$
Show that $\tilde{\phi}$ is bijective, thus an isomorphism (the canonical isomorphism between the two constructions).
This book might help you: John M. Lee. $\textit{Introduction to Smooth Manifolds}$. Springer Science & Business Media, 2nd edition, 2012, p.p. 171-179.
