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I'm referring to Awodey's Category Theory, precisely to the chapter 3.1 on duality principle. Here it is first stated the formal duality: if a statement in the language of category theory can be proved only by the axioms of category theory, also the dual statement (which Awodey specifies formally how to obtain) can be proven, for the axioms are self-dual. Instead the conceptual duality amounts to: if a statement is true interpreted in any category, the dual statement is too. This is justified saying that if a statement is true interpreted in a category $C$, its dual is true interpreted in $C^{op}$ (is this a consequence of the fact that the duality reflects and preserve equality? Is "$=$" the only predicate in the language of categories?); and obviously all the dual categories are just all the categories.

About the relationship between (the "strongness" of) these duality principles, I'd like to know if a statement true in all the categories is provable by the axioms; the converse I'd say is true already. In a certain sense this should be a completeness property, but I'd say that it has not much in common with the first order logic, already by the facts that the structure is made of two sets (objects and arrows) and the composition is defined only on some pairs of arrows for example. Thanks in advance.

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    $\begingroup$ Your question isn't very clear. Equality of arrows is the only predicate symbol in the language of categories. We have a useful word "strength" in English which means what you mean by "strongness". It is unclear what you would "like to know" in the rest of your second paragraph. If you make appropriate allowances for the fact that composition of arrows is a partial operator, the first-order axiomatisation of the notion of category (which is most naturally taken as a two-sorted language with a sort for objects and a sort for arrows) is unaffected if you replace all arrows by their duals. $\endgroup$
    – Rob Arthan
    Commented Jun 16, 2022 at 23:41
  • $\begingroup$ @RobArthan I just wanted to know if the first sentence of the answer below (that surely is phrased better than mine) was indeed true. Btw, "strongness" is a word too $\endgroup$ Commented Jun 17, 2022 at 6:45
  • $\begingroup$ "Strongness" is described by the OED as obsolete except as a nonce-word. I am sorry if it's a bit harder to say, but the English-speaking world has fixed on "strength". $\endgroup$
    – Rob Arthan
    Commented Jun 17, 2022 at 19:42
  • $\begingroup$ It's hard to speak for the whole English-speaking world, as long as the word is normally included in any online dictionary, but I won't insist. $\endgroup$ Commented Jun 17, 2022 at 22:37
  • $\begingroup$ If the OED says a word is obsolete, that's good enough for me. $\endgroup$
    – Rob Arthan
    Commented Jun 18, 2022 at 0:27

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Yes, every first-order statement about categories which is true in all categories (in fact it suffices to consider only small categories) is provable from the category theory axioms, and to deduce this from the usual completeness theorem for first-order logic it suffices to write down a first-order axiomatization of small categories.

This can be done as follows: we think of a small category entirely in terms of its set $\text{Arr}$ of arrows. Objects are encoded as identity arrows which we'll isolate later. Then we introduce two functions $s, t : \text{Arr} \to \text{Arr}$ which send an arrow to its source (identity arrow) and its target (identity arrow); the identity arrows are exactly the arrows which are fixed points of both $s$ and $t$. To encode composition we add a ternary predicate $\{ f \circ g = h \} \subset \text{Arr}^3$ describing which pairs of arrows compose to which. Then we write down a bunch of first-order axioms which can be found here.

I don't have a clear sense of how powerful the first-order language of categories is off the top of my head though. The technical restriction to small categories here, which is probably removable, means this result doesn't directly apply to interesting large categories such as $\text{Set}$ or $\text{Vect}$.

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  • $\begingroup$ Thanks for the link too! I have been searching online for a little introduction to categories in that sense, but I had found nothing. $\endgroup$ Commented Jun 17, 2022 at 7:01

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