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From Set Theory and Linear Algebra, we have these two theorems:

  1. Given two finite sets of the same cardinality $X$ and $Y$ and a function $f:X\rightarrow Y$, the following are equivalent:
    • $f$ is a bijection
    • $f$ is an injection
    • $f$ is a surjection
  2. Given two finite dimensional vector spaces of the same dimension $V$ and $W$ and a linear map $T:V\rightarrow W$, the following are equivalent:
    • $f$ is an isomorphism
    • $f$ is a monomorphism
    • $f$ is an epimorphism

The parallel nature of these theorems suggests they are related. And I have an inkling that Category Theory can make the relationship explicit.

Functors preserves isomorphisms in general, so if I can find a combination of functors which preserve epimorphisms and monomorphisms in a convenient way, we can actually show that the first theorem implies the second.

I know there is an adjoint pair $G\dashv F$ where $F$ is the forgetful functor from Vect $_K$ to Set and $G$ is the functor which sends a set to the free vector space generated by it. And I also know that both $F$ and $G$ are faithful (but not in general full).

But I am having difficulty in getting these facts to work together nicely. Any help in making my ideas rigorous, or an explanation why I'm barking up the wrong tree, is appreciated.

EDIT

If we considered the free v.s. functor from FinSet to FinVect $_K$, we will then have a faithful functor. Since faithful functors reflect monomorphisms and epimorphisms, if we can find a parallel mono (or epic) map that is in the range of the free v.s. functor for every mono (or epic) map we encounter, we'd be able to show the 1st theorem implies the 2nd. Using the fact that vector spaces of the same dimension are isomorphic, we can conclude that the free v.s. functor is also dense. Hence if $V$ and $W$ are vector spaces of the same dimension, and $m:V\hookrightarrow W$ we'd be done if we can find an $f:X\rightarrow Y$ which makes the following diagram commute:

$\hskip2in$ enter image description here

I'm sure a clever choice of bases on $V$ and $W$ and subsequent choices for the isomorphisms would make this possible, but these choices essentially use the 2nd theorem. And this approach doesn't seem to port over to epimorphisms.

  • But a finite dimensional vector space is not a finite set, and not every linear map is induced by a map between the bases... – Zhen Lin Jun 24 '14 at 21:41
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    I'm aware. Those were two reasons I was having difficulty. I'm not aware of an adjoint pair between FinVect $_K$ and FinSet. –  Jun 24 '14 at 21:45
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    I don't believe the two facts are directly related. For instance, if we generalise slightly to modules over a noetherian ring, then finitely generated modules have the property that surjective endomorphisms are isomorphisms, but injective endomorphisms may fail to be isomorphisms. (In other words, f.g. modules over noetherian rings are hopfian.) – Zhen Lin Jun 24 '14 at 21:48
  • Just a thought: maybe there is some connection if you view a pointed finite set as a vector space over the "field with one element". – Santiago Canez Jun 24 '14 at 23:14
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    @ZhenLin: No Noetherian hypothesis is needed - f.g. modules over any commutative ring are Hopfian (see e.g. here) – zcn Jun 25 '14 at 02:57
  • @zcn: Sure, but on the other hand in the statement "noetherian modules are Hopfian" we don't need commutativity. In fact, noetherian objects in abelian categories are Hopfian. But finitely generated modules over non-commutative rings don't have to be Hopfian. – Martin Brandenburg Jun 26 '14 at 13:16
  • The free vector space functor is not dense. It is true that every injective linear map is isomorphic to one in the image of the free vector space functor: use e.g. the Steinitz exchange lemma. The same is true for surjective linear maps: pick a basis for the kernel, extend it to a basis for the whole domain, etc. – Zhen Lin Jun 26 '14 at 14:33
  • Is there not a finite set for every finite cardinality in FinSet? I considered FinVect $_K$ as the codomain of the free v.s. functor. –  Jun 26 '14 at 14:58

1 Answers1

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First of all, there is a purely linguistic connection: An object $A$ of a category is called Hopfian (or co-Hopfian) if every epimorphism (or monomorphism) $A \to A$ is an isomorphism. The Hopfian objects in $\mathsf{Set}$ are precisely the finite sets, the Hopfian objects in $\mathsf{Vect}_K$ are precisely the finite-dimensional vector spaces. Actually these are also the Co-Hopfian objects. In the case of finite-dimensional vector spaces this is because they admit a self-duality, namely the dual vector space. For more examples of Hopfian objects, check out the Wikipedia article.

One might ask if the characterization of Hopfian objects in $\mathsf{Set}$ is equivalent to the one in $\mathsf{Vect}_K$. Here is a proof of a part of $\Rightarrow$:

Let $V$ be a Hopfian vector space. Choose a basis $B$. Then $B$ is a Hopfian set: Every surjection $f : B \to B$ induces a linear surjection $\overline{f} : V \to V$, which must be an isomorphism. In particular it is injective, so that also its restriction $f$ is injective, i.e. $f$ is an isomorphism. Since $B$ is Hopfian, it is finite, and therefore $V$ is finite-dimensional. Conversely, if $V$ is finite-dimensional, then $V$ is Hopfian, but I doubt that this may be reduced to set theory.

A concise proof of this first shows that finite-dimensional vector spaces are Noetherian, and then one shows that Noetherian objects in abelian categories are Hopfian: If $f : A \to A$ is an epimorphism, then $\ker(f) \subseteq \ker(f^2) \subseteq \dotsc$ has to stabilize, say $\ker(f^n)=\ker(f^{n+1})$. Then $\ker(f)=\ker(f) \cap \mathrm{im}(f^n)=f^n(\ker(f^{n+1}))=f^n(\ker(f^n))=0$.

While writing this, I realize that an analogous proof can also be carried out in non-linear categories such as $\mathsf{Set}$ (... details may be be found by the interested reader?).

Martin Brandenburg
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