# Associating to any vector space its $k$-linear dual and the resulting functor

From my understanding, a contravariant functor of a category $$\mathcal{C}$$ can be defined using the notion of opposite category, $$\mathcal{C}^{op}$$. Then for two categories $$\mathcal{C}$$ and $$\mathcal{D}$$, the contravariant functor $$\mathcal{C} \rightarrow \mathcal{D}$$ is the covariant functor $$\mathcal{C}^{op} \rightarrow \mathcal{D}$$.

Now let $$\mathcal{C}$$ be the category of vector spaces $$V$$ over a field $$k$$. Take any vector space $$V$$ and associate to it its $$k$$-linear dual, $$V^{*}$$.

How can I see whether this correspondence results in a covariant or contravariant functor? I think my confusion arises from the fact that both $$\mathcal{C}$$ and $$\mathcal{C}^{op}$$ consist of the same objects. Note that, unfortunately, my understanding of Category Theory is basic.

The main idea is that a contravariant functor reverses the arrows.

As to your concrete example: if $$f:V\to W$$ is a linear transformation, then the dual map $$f^*:W^*\to V^*$$.

Edit. (Some more details.) A covariant functor $$F:\mathscr{C}\to \mathscr{D}$$ is such that $$F(gf)=F(g)F(f)$$ for $$f:X\to Y$$ and $$g:Y\to Z$$.

A contravariant functor $$F':\mathscr{C}\to \mathscr{D}$$ is such that $$F'(gf)=F'(f)F'(g)$$.

In the dual category $$\mathscr{C}^{\text{opp}}$$, we have $$\text{Hom}_{\mathscr{C}^{\text{opp}}}(X,Y):=\text{Hom}_{\mathscr{C}}(Y,X)$$ and composition is done 'the other way around':

$$\text{Hom}_{\mathscr{C}^{\text{opp}}}(X,Y)\times \text{Hom}_{\mathscr{C}^{\text{opp}}}(Y,Z)\to \text{Hom}_{\mathscr{C}^{\text{opp}}}(X,Z),(f,g)\mapsto fg.$$

Hence, we can reformulate by saying that $$F'$$ is a covariant functor $$\mathscr{C}^{opp}\to \mathscr{D}$$

As to the example of the dual vector space, we take

$$f:V\to W,g:W\to Z,$$ then

$$Ff:W^*\to V^*,Fg:Z^*\to W^*$$ and

$$F(gf):Z^*\to V^*,$$

so $$F(gf)=F(f)F(g)$$. This is how we see that $$(-)^*$$ is contravariant.

• I understand that the idea is to reverse the arrows, what I am finding confusing is following: if I suggest an association between the objects then such association goes both ways, and so there is no clear way by which I can say which direction will correspond to the covariant functor and which to the contravariant. I am not sure I have made it clearer Apr 24 '20 at 16:05
• @Morettin I am expanding my answer a bit to make it clearer :) Apr 24 '20 at 16:06
• Thank you! So would it be right to say that if I were to take $\mathcal{C}$ = $\mathcal{C}^{op}$ then by the same process we would have a covariant functor? And so, in some sense, whether a functor is covariant or contravariant is strongly dependent on what our choice of category is? Apr 24 '20 at 16:29
• No, I don't think we could say that.. A functor $F:\mathscr{C}\to \mathscr{D}$ that sends $f:X\to Y$ to $Ff:FX\rightarrow FY$ is called contravariant; a functor $F:\mathscr{C}\to \mathscr{D}$ that sends $f:X\to Y$ to $Ff:FX\leftarrow FY$ is called contravariant. You should not complicate things :P Apr 24 '20 at 16:32