Ambiguous adjoint awareness

Question

So I've seen the demonstration of the free functor giving all linear combos being left adjoint to the forgetful functor. However I can't convince myself that the inverse isn't true using explicit construction of an isomorphism.

Take free functor F and forgetful functor U moving between Vectk and Set for some field k. If presuming a linear map ${g}: F(S)\rightarrow V$, where S is a set and V is a vectorspace, we can define $\bar{g}:S \rightarrow U(V)$ by $\bar{g}(s)=g(s)$ for $s\in S$, Why can't we also have the 'inverse' defined $\bar{g}(U(v))=g(v)$ for $v\in V$.

Similarly with an $f:U(V)\rightarrow S$ why can one not have a function between set and Vectk defined by $\bar{f}:V\rightarrow F(S)$ defined by $\bar{f}(\sum_{e\in U(V)}\lambda_{e}v=\sum_{e\in U(V)}\lambda_{e}f(v)$. Where here $v \in U(V)$.

While this looks odd I can't see why it doesn't give the free functor as the right adjoint of the forgetful. What I'm looking for here is to appreciate what it is I'm not quite understanding about the situation ;as write I get the distinct impression I have some fundamental misapprehension I'm not able to identify.

Answer 1

First: because $g(v)$ is not generally an element of $S$. Second: this only makes sense if you've chosen a basis. Your $f$ already has a value chosen for every element of $V$, so it would be as sensible to set $\bar f=f$; but this will never be linear.

Answer 2

I don't understand your construction so I can't tell you where your construction failS.

Nonetheless I'll tell why your construction cannot work.

By general results on adjoint functors a left adjoint preserves colimits, so if $U$ was a left adjoint it should preserve colimits, hence coproducts.

Now let consider the finite field $\mathbb F_2$ of two elements. We know that $\mathbb F_2\oplus \mathbb F_2\oplus \mathbb F_2\cong \mathbb F_2^3$, so we should have that $U(\mathbb F_2^3)\cong 3U(\mathbb F_2)$ hence the set $U(\mathbb F_2^3)$ should have cardinality 6, which is absurd.