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.

  • $\begingroup$ 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 $\endgroup$
    – Morettin
    Apr 24 '20 at 16:05
  • $\begingroup$ @Morettin I am expanding my answer a bit to make it clearer :) $\endgroup$
    – rae306
    Apr 24 '20 at 16:06
  • $\begingroup$ 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? $\endgroup$
    – Morettin
    Apr 24 '20 at 16:29
  • $\begingroup$ 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 $\endgroup$
    – rae306
    Apr 24 '20 at 16:32

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