Universal Property of free vector space

Question

Let $X$ be any set and $F$ a field. Let $V_{X}$ be set of functions $f:X \to F$ for which $f(x)$ is nonzero for finitely many $x \in X$. Note that $V_{X}$ is a vector space.

Call $\sigma_{x}$ a function such that $\sigma_{x}(x) = 1$ and $\sigma_{x}(y) = 0$ for each $x \in X$ and $y \in X-\{0\}$. Call $\gamma$ another function such that $\gamma: X \to V_{X}$ such that $\gamma(x) = \sigma_{x}$. Prove that if $\phi: X \to V$ is any function to a vector space over $F$, then there is a unique linear map $\bar\phi: V_{X} \to V$ such that $\phi = \bar\phi \circ \gamma$

I was told that what I need to prove is called universal property, and it seemed to be related to a lot with tensor products, but I am not sure if this is the same proof as that. Can someone help me with how to approach this? Thank you.

Answer 1

Let $\phi:X\rightarrow V$ be any function. Let $f\in V_X$, that is, let $f:X\rightarrow F$ be a map such that $f(x)\neq 0$ for only finitely many values of $x$.

We claim that $f=\sum_{x\in X}f(x)\delta_x$. Indeed, let $y\in X$, then $\sum_{x\in X}f(x)\delta_x(y)=f(y)\delta_y(y)=f(y)$. Also note that $\sum_{x\in X}f(x)\delta_x$ is actually a finite linear combination of the $\delta_x$'s as $f(x)$ is non-zero for only finitely many $x$.

We need to define a linear map $\overline{\phi}$ extending $\phi$. By the above, we have that \begin{eqnarray} \overline{\phi}(f)&=&\overline{\phi}(\sum_{x\in X}f(x)\delta_x)\\ &=&\sum_{x\in X}f(x)\overline{\phi}(\delta_x)\\ &=&\sum_{x\in X}f(x)\overline{\phi}(\gamma(x))\\ &=&\sum_{x\in X}f(x)\phi(x).\\ \end{eqnarray} Hence $\overline{\phi}$ is completely determined.

All of this is just a fancy way of saying that you are considering a $F$-vector space with basis $X$. This has nothing to do with tensor products (so don't go into that right now). You can indeed use "universal properties" to define certain things and what we did is an example of a universal property that characterizes "freeness" in some sense. I don't believe this example is a good way to introduce universal properties to someone, so unless you are required to understand universal properties right now, I wouldn't bother too much with it.

Answer 2

It has been my experience that the free space $F<X>$ is an excellent way to introduce universal properties. It is easy to state and prove the universal property of free spaces and also easy to deduce several standard corollaries of universal properties.

The more general free space $V<X>$ where $V$ is a vector space over a field $F$ is naturally isomorphic to the tensor product $F<X> \otimes V$, so tensor products are very much involved with free vector spaces.