# Which (endo)functors of the category of finite-dimensional real vector spaces induce continuous maps between Hom-sets?

Let $\operatorname{Vect-fin}$ be a category of finite-dimensional vector spaces over $\mathbb{R}$. In this category Hom-sets $\operatorname{Hom}(V,W)$ are themselves finite-dimensional vector spaces over $\mathbb{R}$. Now let $F$ be a functor from $\operatorname{Vect-fin}$ to $\operatorname{Vect-fin}$.

My question is what we have to assume on $F$ to conclude that induced maps: $$\tag{#} \operatorname{Hom}(V,W)\xrightarrow{F}\operatorname{Hom}(F(V),F(W))$$ are continuous for all $V,W\in \operatorname{Vect-fin}$?
(Recall that any finitely dimensional vector space over $\mathbb{R}$ can be equipped with unique topology with respect to which algebraic operations are continuous - we consider this topology on Hom-sets.)

Motivation

When I learnt for the first time about topological vector bundles, I saw that the usual constructions (better: functors) which are available for vector spaces: direct sum, tensor product, exterior and symmetric powers, dual (and more generally $\operatorname{Hom}$), are also doable in the category of vector bundles over fixed topological space.

The argument can be roughly arranged as follows:

1. for given vector bundles take a common trivialization cover of our space,
2. take transition maps,
3. apply our functor from the category of vector spaces "point by point" to the transition maps,
4. rejoice in the fact that maps thus obtained are continuous, giving transition maps for a new vector bundle, which over our trivialization cover looks like trivial bundle with fiber equal to what we should get when we apply our functor in the category of vector spaces to the fibers of given vector bundles.

If one write down the details, it turns out that we are composing (the tuples of) the continuous maps $U\to \operatorname{GL}_{n}(\mathbb{R}) \subset \operatorname{Hom}(\mathbb{R}^{n},\mathbb{R}^{n})$ with map analogous to (#) above (it may look more general if our functor goes from some cartesian power of $\operatorname{Vect-fin}$, but for $F =\Lambda^{n}$ it is exactly the map (#)), and what makes us successful and happy is that this composition is still continuous. (Moreover analogous construction works in the smooth or complex-analytic category.)
Thus the proof works equally well for all functors for which the induced map between Hom-sets is continuous.
Despite the fact that I have never worked with quite general functors between categories of vector spaces I believe that the answer could be conceptually enlightening or at least pleasant.

Starting points

Perhaps we should begin with the question about the existence of functor such that (#) is not continuous. Another observation - if the additive functor preserve scalar multiplication in Hom-sets, then (#) is continuous, as linear map of finitely dimensional vector spaces. But what happens for general additive functors? Are there any which do not preserve scalar-multiplicative structure?

(One closing remark: I am likewise eager to know what is going on in smooth and complex-analytic categories. And if you still strive for generalizations, you are welcome to replace $\mathbb{R}$ by any topological field, or even topological ring and consider modules there).

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The generalisation cannot hold true in general: if we replace $\mathbb{R}$ by $\mathbb{C}$, then we can choose a discontinuous ring automorphism $\sigma : \mathbb{C} \to \mathbb{C}$, which will then induce an additive automorphism (!) of $\textbf{FinVect}_\mathbb{C}$ that acts discontinuously on hom-sets. – Zhen Lin Nov 10 '12 at 22:30
@Zhen Lin: Could you elaborate a little how the induced additive functor is defined? In particular, don't you need to fix a base on each vector space in some canonical way? – pluripolar bear Nov 11 '12 at 12:10
The obvious way of doing it would be to fix a basis for each vector space, yes. But you can show this still gives a functor. (A simple example: if we take $\sigma : \mathbb{C} \to \mathbb{C}$ to be complex conjugation, then the functor I am thinking of is the one that conjugates all the entries of a matrix.) – Zhen Lin Nov 11 '12 at 13:00
Or in other words you fix an ("almost linear") action of $\operatorname{Aut}(\mathbb{C})$ on each $V\in \operatorname{FinVect}$ and send each $f\in \operatorname{Hom}(V,W)$ to $\sigma f\sigma^{-1}\in \operatorname{Hom}(V,W)$ for fixed discontinuous $\sigma\in \operatorname{Aut}(\mathbb{C})$. Strangely, I had a problem with convincing myself that arbitrary choice of basis / action would do, but now it seems clear to me. Thanks. – pluripolar bear Nov 11 '12 at 13:52