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The well-known duality principle asserts that if a statement is true (resp. false) for a category $\mathfrak{C}$, then the dual statement must be also true (resp. false) for the dual category $\mathfrak{C}^{\mathrm{op}}$. Now, in some cases it is useful to work with the dual category because it might simplify the analysis. However, in this case, by the duality principle, we translate the found properties with respect to $\mathfrak{C}$. Hence, my quedtion is: except for what I remarked before, why it is so important to study dual categories? Honestly, I considered such a notion useful only to simplify the work needed when using $\mathfrak{C}$.

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    $\begingroup$ It simplifies statements and people who don't like to work contravariant can work covariant. $\endgroup$
    – user943633
    Jul 10, 2021 at 15:35
  • $\begingroup$ So, at an interpretative level, they do not give new informations than the starting category. I think for instance to frames and locales. Why even changing names, if opposite only gives a formal way to simplify computations ? $\endgroup$ Jul 10, 2021 at 15:42
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    $\begingroup$ The difference is intuition. Frames are like algebras/lattices and locales are like spaces. $\endgroup$ Jul 10, 2021 at 17:27
  • $\begingroup$ In a certain sense, this follows because taking locales is a generalization of taking open subsets with inclusions, while taking frames underline only the lattice-structure of open sets, right? Therefore, there exists in general an interpretative reason to introduce opposite category (they give a different point of view for seeing mathematical structures). Hence, in particular, opposites must not be seen only as a "formal convenience" to find further properties of $\mathfrak{C}$. Is it right? $\endgroup$ Jul 10, 2021 at 21:01

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What is important is that dual categories appear naturally, most notably because there are (what people call) "contravariant" functors - what I prefer to call functors on dual categories (it is very convenient to declare all functors as "covariant"). A basic example is the functor $\mathbf{Top}^{\mathrm{op}} \to \mathbf{CRing}$, $X \mapsto C(X)$, $f \mapsto f^*$, a more complicated example is singular cohomology $H^* : \mathbf{Top}^{\mathrm{op}} \to \mathbf{grAb}$. Also, every (pre)sheaf on a space $(X,\mathcal{O})$ is actually a functor on $\mathcal{O}^{\mathrm{op}}$.

One important observation is also that dual categories of concrete categories in practice behave quite differently than these concrete categories. For example, although the dual of every abelian category is also abelian, the dual of a Grothendieck category is almost never a Grothendieck category, which applies in particular to $(\mathbf{Mod}_R)^{\mathrm{op}}$, the dual of the category of right $R$-modules, where $R$ is a nontrivial ring.

Also, concrete representations of these dual categories often involve some kind of topology. In fact, Pontryagin duality tells us (in a special case) that there is an equivalence of categories between $\mathbf{Ab}^{\mathrm{op}}$ and the category $\mathbf{CompAb}$ of compact topological abelian groups. It follows formally that $(\mathbf{Mod}_R)^{\mathrm{op}}$ is equivalent to the category ${}_R \mathbf{CompMod}$ of compact topological left $R$-modules.

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  • $\begingroup$ PS: I just babbled a bit about why I find dual categories important. I have no idea if this answers the given question. $\endgroup$ Jul 10, 2021 at 22:40
  • $\begingroup$ Very good answer! Following your examples, I have another question, which is related to the first I posed: the consequences of the equivalence between ${\bf Ab^{op}}$ and ${\bf CompAb}$ may be translatable as properties of the category ${\bf Ab}$? Indeed, if each property of dual category may be translated as a property of ${\bf Ab}$, then the introduction of dual category, beyond giving a formal tool in order to see in a simpler way further properties of the starting category, has no very importance. Namely, I can study a category directly, using contravariance instead of passing to dual. $\endgroup$ Jul 11, 2021 at 13:14
  • $\begingroup$ I don't understand what you specifically want to know. $\endgroup$ Jul 11, 2021 at 13:41
  • $\begingroup$ Does the previous equivalence provide properties of the opposite ${\bf Ab}$ which cannot be translated using duality into properties of ${\bf Ab}$? $\endgroup$ Jul 11, 2021 at 14:24

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