Duality of linear independence and span I have observed the following phenomenon in linear algebra.
Let $V$,$W$ be two vector spaces over a field $F$. Let $T:V\to W$ be a linear map.
On object level, we have
If $S'\subseteq S\subseteq V$ subsets, then

*

*$S$ is a linearly independent set $\implies$ $S'$ is a linearly independent set.

*$S'$ is a spanning set $\implies$ $S$ is a spanning set.

On morphism level, we have
If $S\subseteq V$ is a subset, then

*

*If $T$ is injective on $S$, then $T(S)$ is a linearly independent set $\implies$ $S$ is a linearly independent set.

*If $T$ is surjective, then $S$ is a spanning set $\implies$ $T(S)$ is a spanning set.

I have a strong feeling linear independence is "dual" to span, but cannot find a categorical formulation about this anywhere. Is there a systematic way to formulate this phenomenon in category theory?
Thanks a lot in advance.
 A: The phenomenon can be phrased in terms of free objects. It might have been hard to see how this phenomenon can be phrased categorically due to there being two categories at play here (namely, the category of vector spaces, and the category of sets).
More precisely, let $\def\Vect{\mathbf{Vect}}\Vect$ be the category of vector spaces and linear transformations (over a fixed base field), and $\def\Set{\mathbf{Set}}\Set$ the category of sets and functions, and take $U:\Vect\to\Set$ to be the forgetful functor that sends a vector space to its underlying set. This functor admits a left adjoint $F:\Set\to\Vect$ which sends a set $S$ to the "free vector space" generated by the set; i.e., the vector space obtained from taking the elements of $S$ to be a basis.
The universal property of these free objects is simply:

There is a correspondence between linear transformations $F(S)\to W$ and functions $S\to U(W)$. In other words, linear transformations $F(S)$ are completely determined by where it sends its basis $S$.

Now, let $S\subseteq V$ be a subset of your vector space. Then, more properly, this means you are taking an inclusion $S\hookrightarrow U(V)$ into the underlying set of $V$. We can use the left adjoint $F$ to characterise "linearly independent set" and "spanning set:"

*

*$S$ is linearly independent if and only if the induced linear transformation $F(S)\to V$ is injective (that is, a monomorphism in $\Vect$)

*$S$ is a spanning set if and only if the induced linear transformation $F(S)\to V$ is surjective (that is, an epimorphism in $\Vect$)

If you look carefully, these statements say the exact same thing as in the usual linear-algebraic definitions of these terms.
In this language, the duality is much more apparent: monomorphisms and epimorphisms are formally dual in any category (in the sense that a morphism in $\mathcal C$ is a monomorphism iff its corresponding morphism in $\mathcal C^{\mathrm{op}}$ is an epimorphism).

If you like, you can also try to prove your observations in this language.
For instance, let $S'\subseteq S\subseteq U(V)$.
Denote the inclusion of subsets by $i:S'\hookrightarrow S$.
Then, the induced map $F(S')\to V$ is the composite $F(S')\xrightarrow{Fi}F(S)\to V$.

*

*If $F(S')\to V$ is an epimorphism, then $F(S)\to V$ must be an epimorphism as well. In other words, if $S'$ is a spanning set, then so is $S$.

*You can check that $Fi:F(S')\to F(S)$ is a monomorphism.
Therefore, if $F(S)\to V$ is a monomorphism, so is the composite $F(S')\to V$. In other words, if $S$ is linearly independent, then so is $S'$.


A word of warning: $Fi$ being a monomorphism does not just follow from $F$ being a left adjoint---generally, left adjoints need not preserve monomorphisms.
This is an instance of a more general phenomenon, though: you could say this happens because the forgetful functor $U$ is faithful and maps to $\Set$, as explained in this answer.

Edit: The second part is harder, because there is a nuance: if $T$ maps multiple vectors in $S$ to the same vector in $T(S)$, then you have to deem $T(S)$ linearly dependent (so in some sense, you are implicitly viewing $T(S)$ as a list, or a multiset, rather than a mere set. (This was pointed out by @Thinknot; I made the same oversight.)
While one could work to make your observation categorically precise, instead there is probably a more satisfactory "duality" that looks similar:

*

*If $T:V\to W$ is injective, and $T(S)$ is linearly independent in $W$, then $S$ is linearly independent in $V$

*If $T:V\to W$ is surjective, and $S$ is a spanning set of $V$, then $T(S)$ is a spanning set of $W$
To see (categorically) why these are true, note we have this commutative square
$\require{AMScd}$
\begin{CD}
F(S) @>>> V \\
@VFTVV @VVTV \\
F(TS) @>>> W
\end{CD}
To see (1): if $T$ is injective, then so is its restriction $S\to T(S)$, which means that both vertical arrows in the above square are monic.
If $T(S)$ is linearly independent, then this means $F(TS)\to W$ is monic as well, meaning the composite $F(S)\to V\to W$ is monic, and thus that $F(S)\to V$ is monic. In other words, $S$ is linearly independent in $V$.
To see (2): if $T$ is surjective, and $S$ is a spanning set, then $F(S)\to V\to W$ is a composite of epis, and is thus epi. This makes $F(S)\to F(TS)\to W$ an epi, implying $F(TS)\to W$ is an epi. Notice that these arguments seem more like formal duals of each other.
