What is Span? Why should we give up the traditional definition?

Let $V$$V$ be a vector space over $\mathbb{F}$$\mathbb F$ and $X\subseteq V$$X\subseteq V$. We naturally have an inclusion map $f:X\hookrightarrow V$$f:X\hookrightarrow V$. By the universal property of free vector space, we have the following diagram.

Let us define $\tilde{f}$$\tilde{f}$​ to be the span map.

It is natural since $X$$X$ is linearly independent iff $\tilde{f}$$\tilde{f}$ is injective, $X$$X$ spans $V$$V$ iff $\tilde{f}$$\tilde{f}$ is surjective, and $X$$X$ is a basis iff $\tilde{f}$$\tilde{f}$ is an isomorphism.

Lots of people are confused about why we only allowed finite linear combination even when $X$$X$ is an infinite set.

The reason is simple, $F(X)\cong {\mathbb{F}}^{\oplus X}$$F(X)\cong \mathbb F^{\oplus X}$, which is direct sum, not product. And it is the domain of $\tilde{f}$$\tilde{f}$.

Ok, so what is the connection between the new definition and the old? Well the old one is simply $\mathrm{Im}\tilde{f}$$\mathrm{Im}\tilde{f}$. I would like to say that it is a ridiculous definition, the only thing it did was make people feel confused.

The story is the same for the group as well. Let $X\subseteq G$$X\subseteq G$ and $F(X)$$F(X)$ be the free group generated by $X$$X$, then by the universal property of free group, we have the following diagram commute:

Where $\begin{array}{c}F(X)\cong \ast x_{\in X}\mathbb{Z}\end{array}$$F(X)\cong *x_{\in X}\mathbb Z\\$.

Again, $X$$X$ generates $G$$G$ iff $\tilde{f}$$\tilde{f}$ is surjective.

A really good question is do we have an analogy to linearly independent in $\mathrm{Grp}$$\mathrm{Grp}$?

The answer will not be quite the same. Since every vector space is free but not every group is. But we could capture this idea via $\tilde{f}$$\tilde{f}$ !

Recall that $X$$X$ is linearly independent iff $\tilde{f}$$\tilde{f}$ is injective, so the proper analogy is $\tilde{f}$$\tilde{f}$ is injective, i.e. $\ker (\tilde{f})=\{e\}$$\mathrm{ker}(\tilde f)=\{e\}$ or $⟨X⟩$$\langle X\rangle$ is a free group.

Indeed, the $\ker (\tilde{f})$$\mathrm{ker}(\tilde{f})$ gives us a way to measure how efficient the generator set we use. Since the kernel represents what we lose. So if $\ker ({\tilde{f}}_X)\le \ker ({\tilde{f}}_Y)$$\mathrm{ker}(\tilde f_X)\le\mathrm{ker}(\tilde{f}_Y)$, then $X$$X$ is a more efficient generator set than $Y$$Y$. Here the notation $\le$$\le$ means subobject.

i.e. $G\le G^{\prime }$$G\le G'$ iff there exists a monomorphism $\iota :G\to G^{\prime }$$\iota:G\to G'$ in $\mathrm{Grp}$$\mathrm{Grp}$.

The reason that we do not use $\subseteq$$\subseteq$ is for $X\ne Y,F(X)\ne F(Y)$$X\ne Y,F(X)\ne F(Y)$.

Now let us back to category of vector space over $\mathbb{F}$$\mathbb F$.

Easy to see that

$$\begin{array}{}\text{(1)}& \ker ({\tilde{f}}_X)\le \ker ({\tilde{f}}_Y)⟺|X|\le |Y|\end{array}$$

Hence we only need to use the card of the generator set to measure it is efficient or not in Category of vector space. But things are different in category of group.