Exercise 12.6 (Characteristic Property of the Free Vector Space)

Question

extbf{Proposition 12.5 (Characteristic Property of the Free Vector Space).} For any set $S$ and any vector space $W$, every map $A: S 	o W$ has a unique extension to a linear map $\widebar{A}: \mathcal{F}(S) 	o W$.

Elu Thingol · Accepted Answer

\begin{proof}
Since $f \in \mathcal{F}(S)$ model linear combinations, we should try and treat it as such in the definition we are about to give for $\overline{A}$. First notice, that since $f$ is a replacement for finite sums, we can always enumerate the associated values $x_1^{(f)},\ldots,x_{n_{f}}^{(f)} \in S$ for which $f(x_i) 
eq 0$ by a finite index $i=1,\ldots,n_f$. Using this, we rewrite $f = \sum_{i=1}^{n_1}f(x_i)\cdot \delta_x$ and define
\begin{align*}
    \overline{A}:\mathcal{F}(S) 	o W, \qquad \overline{A}(f) &= \overline{A}\left(\sum_{i=1}^{n_1}f(x_i)\cdot \delta_xight) 	extcolor{gray}{\stackrel{	ext{should be}}{=} \sum_{i=1}^{n_f} f(x_i) \cdot \overline{A}(\delta_x)}\
    &:= \sum_{i=1}^{n_1} f(x_i) \cdot A(\delta_x).
\end{align*}

We verify that the given definition yields the desired properties.
\begin{itemize}
    \item 	extit{(extension)} Our extended function $\overline{A}$ should agree with $A$ on the common domain (where we use that natural identification $x \in S \longleftrightarrow \delta_x \in \mathcal{F}(S)$). We then trivially obtain from the definition:
    \begin{equation*}
        \overline{A}(\delta_x) = A(x).
    \end{equation*}
    \item 	extit{(uniqueness)} This is obvious from the definition of $\overline{A}$.
\end{itemize}
\end{proof}

Exercise 12.6 (Characteristic Property of the Free Vector Space)

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Exercise 12.6 (Characteristic Property of the Free Vector Space)

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