free vector space over a set

In this entry we constructthe free vector space over a set, orthe vector space generated by a set [1].For a set $X$, we shall denote this vector space by $C(X)$.One application of this construction is given in [2],where the free vector space is used to define the tensor product formodules.

To define the vector space $C(X)$, let us first define $C(X)$ as aset. For a set $X$ and a field $𝕂$, we define

$C(X)$$=$$\mathrm{\{}f:X\to 𝕂|f^{-1}(𝕂\backslash \backslash \{0\})\text{is finite}\}.$

In other words, $C(X)$ consists of functions $f:X\to 𝕂$that are non-zero onlyat finitely many points in $X$.Here, we denote the identity element in $𝕂$ by $1$, andthe zero element by $0$.The vector space structure for $C(X)$is defined as follows. If $f$ and $g$ arefunctions in $C(X)$, then $f+g$ is the mapping $x\mapsto f(x)+g(x)$.Similarly, if $f\in C(X)$ and $\alpha \in 𝕂$, then$\alpha f$ is the mapping $x\mapsto \alpha f(x)$. It is not difficult tosee that these operations are well defined, i.e., both $f+g$ and$\alpha f$ are again functions in $C(X)$.

0.0.1 Basis for $C(X)$

If $a\in X$,let us define the function ${\mathrm{\Delta }}_a\in C(X)$ by

${\mathrm{\Delta }}_a(x)$$=$

These functions form a linearly independent basis for $C(X)$, i.e.,

$C(X)$$=$$span{\{{\mathrm{\Delta }}_a\}}_{a\in X}.$(1)

Here, the space $span{\{{\mathrm{\Delta }}_a\}}_{a\in X}$ consists of allfinite linear combinations of elements in ${\{{\mathrm{\Delta }}_a\}}_{a\in X}$.It is clear that any element in $span{\{{\mathrm{\Delta }}_a\}}_{a\in X}$is a member in $C(X)$.Let us check the other direction. Suppose $f$ is a member in $C(X)$.Then, let${\xi }_1,\mathrm{\dots },Ã\mathrm{‚}Â\mathrm{ }{\xi }_N$ be the distinct points in $X$ where $f$ is non-zero.We then have

$f$$=$${\displaystyle \sum _{i=1}^N}f({\xi }_i){\mathrm{\Delta }}_{{\xi }_i},$

and we have established equality in equation 1.

To see that the set ${\{{\mathrm{\Delta }}_a\}}_{a\in X}$ is linearly independent,we need to show that its any finite subset is linearly independent.Let $\{{\mathrm{\Delta }}_{{\xi }_1},\mathrm{\dots },{\mathrm{\Delta }}_{{\xi }_N}\}$ be sucha finite subset, andsuppose ${\sum }_{i=1}^N{\alpha }_i{\mathrm{\Delta }}_{{\xi }_i}=0$ for some${\alpha }_i\in 𝕂$. Since the points ${\xi }_i$ are pairwise distinct, itfollows that ${\alpha }_i=0$ for all $i$. This shows that the set${\{{\mathrm{\Delta }}_a\}}_{a\in X}$ is linearly independent.

Let us define the mapping $\iota :X\to C(X)$, $x\mapsto {\mathrm{\Delta }}_x$.This mapping gives a bijection between $X$ and the basisvectors ${\{{\mathrm{\Delta }}_a\}}_{a\in X}$. We can thus identify thesespaces. Then $X$ becomes a linearly independent basis for $C(X)$.

0.0.2 Universal property of $\iota :X\to C(X)$

The mapping $\iota :X\to C(X)$ is universal inthe following sense. If $\varphi$ is an arbitrary mapping from $X$ to avector space $V$, then there exists a unique mapping $\overline{\varphi }$such that the below diagram commutes:

Proof. We define $\overline{\varphi }$ as the linear mapping thatmaps the basis elements of $C(X)$ as $\overline{\varphi }({\mathrm{\Delta }}_x)=\varphi (x)$.Then, by definition, $\overline{\varphi }$ is linear. For uniqueness,suppose that there are linear mappings$\overline{\varphi },\overline{\sigma }:C(X)\to V$such that $\varphi =\overline{\varphi }\circ \iota =\overline{\sigma }\circ \iota$.For all $x\in X$, we then have $\overline{\varphi }({\mathrm{\Delta }}_x)=\overline{\sigma }({\mathrm{\Delta }}_x)$.Thus $\overline{\varphi }=\overline{\sigma }$ since both mappings are linear andthe coincide on the basis elements.$\mathrm{\square }$