What really are the elements of the free vector space?

Question

One of the definition of the tensor product $V\otimes W$ is the Quotient space of the free vector space $V * W$ by some ideal $I$ which is spanned by the bilinearity relations. I've seen $V * W$ defined as a span of the cartesian tuple $(v, w)$ where $v \in V$ and $w \in W$, but as far as I know, vector spaces represented as spans are a subspace of a vector space containing those spanning vectors, so it seems circular to describe a vector space as such. For example, if I defined the Euclidean space as $\text{span} \{v : v \in \mathbb{R}^n\}$, then that would be circular, and that seems to be what's happening with this definition of the free space.

Answer 1

The simplest way is to start with a set $X$ and consider the vector space $$\mathbb{R}^X = \{f\colon X\to\mathbb{R}\}$$ of all functions from $X$ to $\mathbb{R}$, which is a vector space over $\mathbb{R}$ with pointwise addition and scalar multiplication (you can replace $R$ with any field).

For each $x$ in $X$, consider the function $\mathbf{x}$ given by $$\mathbf{x}(y) = \left\{\begin{array}{ll} 1 & \text{if }y=x,\\ 0 & \text{otherwise.} \end{array}\right.$$ We can now consider the span of $\{\mathbf{x}\mid x\in X\}$ inside $\mathbb{R}^X$. This is usually denoted by $\mathbb{R}^{(X)}$, and it consists of the functions $X\to \mathbb{R}$ of finite support.

This vector space is the free ($\mathbb{R}$-)vector space on $X$.

So for $V*W$, take $X=V\times W$. Even though $V\times W$ has a structure of a vector space, we are not thinking of it as a vector space, we are thinking of it simply as a set. The vector space $V*W$ is none other than $\mathbb{R}^{(V\times W)}$.

Then we "forget" to use boldface to denote the function, and simply call the function $\mathbf{(v,w)}$ that we defined before as "$(v,w)$". Then elements of $V*W$ will consist of functions ("expressions") of the form $$\alpha_1(v_1,w_1) + \alpha_2(v_2,w_2) + \cdots + \alpha_n(v_n,w_n),$$ where the $(v_i,w_i)$ are pairwise distinct elements of the set $V\times W$; they may repeat, so we could have $v_1=v_2$ and $w_1\neq w_2$, etc.

So we can "think" of it as simply being linear combinations of the elements of the set $V\times W$, i.e., "the span" of the set $V\times W$, and think of that set, via the boldfaced elements, as "the" distinguished basis of the vector space.

What "are" these element when it is at home? Well, in this construction, the element $$\alpha_1(v_1,w_1)+\cdots + \alpha_n(v_n,w_n)$$ is the function whose value at an element $(v,w)$ of $V\times W$ is $\alpha_k$ if $(v,w)=(v_k,w_k)$, and zero otherwise. But what matters isn't what the elements are when they are at home. What matters is that this vector space has the usual universal property associated to a basis: given any vector space $Z$ and any function $g$ from the set $V\times W$ to $Z$, there is one and only one linear transformation from $\mathbb{R}^{(V\times W)}$ to $Z$ that extends $g$.