Timeline for What is the definition of "canonical"?
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21 events
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| Jul 19, 2018 at 10:06 | comment | added | Qfwfq | I realize now you're obviously right: $i$ is indeed not a functor $C\to C^{op}$. I thought there ought to be a well known structure under which my definition (the one that makes sense) was a particular case of: thanks for the info! | |
| Jul 18, 2018 at 23:57 | comment | added | Dmitri Pavlov | @Qfwfq: f↦f^op does not define a functor C→C^op (but rather only a functor C^op→C^op). Your last definition is different, however, and is known under the name of a dinatural transformation: ncatlab.org/nlab/show/dinatural+transformation | |
| Jul 18, 2018 at 20:35 | comment | added | Qfwfq | If you want to spell out "my" definition of $\alpha:F\Rightarrow G$ where $F$ is covariant and $G$ contravariant, without mentioning $i$, you can equivalently say that you give a collection $\alpha_x$ for $x\in\mathrm{ob}(\mathcal{C})=\mathrm{ob}(\mathcal{C}^{op})$ such that, for every $f:x\to y$ in $\mathcal{C}$, $G(f)\circ\alpha_{y}\circ F(f)=\alpha_x$. | |
| Jul 18, 2018 at 20:30 | comment | added | Qfwfq | My $i:\mathcal{C}\to\mathcal{C}^{op}$ was not $f\mapsto f^{-1}$, but $f\mapsto f^{op}$, where, for $f: x\to y$ in $\mathcal C$, $f^{op}$ is the same map but thought of as $y\to x$ in $\mathcal{C}^{op}$. | |
| Jul 18, 2018 at 16:53 | comment | added | Dmitri Pavlov | @Qfwfq: The functor i does not always exist: in the category {0→1} there is a morphism 0→1, but no morphism 1→0. However, i always exists in a groupoid, where we can take f↦f^{−1}. This is exactly what I did above. | |
| Jul 18, 2018 at 9:57 | comment | added | Qfwfq | Maybe there is a way of defining what a natural transformation between a covariant and a contravariant functor (not just their restrictions to a subcategory) should be. Consider $F:\mathcal C \to \mathcal D$ and $G:\mathcal{C}^{op}\to \mathcal{D}$. Then you can say a natural transformation between $F$ and $G$ is a natural transformation (in the usual sense) between $F$ and $G\circ i$ where $i:\mathcal C \to \mathcal{C}^{op}$ is the "formal opposite" $i:x\mapsto x$, $i: (f:x\to y)\mapsto (f^{op}: y\to x)$. | |
| Jul 18, 2018 at 9:46 | comment | added | Qfwfq | @Dmitri Pavlov: well, okay, that's not the usual dualization functor $f^{*}:Y^* \to X^*$, $f^{*}(\varphi):=\varphi \circ f$. | |
| Jul 18, 2018 at 2:28 | comment | added | Dmitri Pavlov | @Qfwfq: Both functors id and * are covariant endofunctors on the category of vector spaces and isomorphisms of vector spaces: id(X)=X, id(f)=f, X*=Hom(X,k), f*=Hom(f,k)^{−1}. | |
| Jul 17, 2018 at 18:18 | comment | added | Qfwfq | I'm late to the party but: how can the identity functor possibly be (naturally or otherwise) isomorphic to the dualization functor?! The identity is covariant while the dualization is contravariant! | |
| Apr 11, 2010 at 5:05 | comment | added | Dmitri Pavlov | @Chris: Well, usually we want to keep all isomorphisms in the subcategory, but your question still stands, because some categories only have identity morphisms as isomorphisms. In this case I am tempted to use the 2-functoriality condition in the second paragraph of my answer (alternatively, disallow the axiom of choice). For example, using the axiom of choice we can construct an algebraic closure functor on the category of fields with identity maps as morphisms, but I do not think there is a way to make this functor depend 2-functorially on the underlying elementary topos. | |
| Apr 9, 2010 at 11:35 | comment | added | Chris Schommer-Pries | @Dmitri: Pass to the subcategory that has the same objects but only has identity morphisms. Then any assignment is functorial. Would you then consider any such assignment "canonical"? | |
| Mar 29, 2010 at 18:03 | comment | added | Dmitri Pavlov | @Jan: The automorphism group of an object is functorial with respect to the subcategory consisting of isomorphisms in the original category. Therefore it is canonical. Notice that sometimes we need to pass to a different category to achieve functoriality. | |
| Mar 29, 2010 at 10:12 | comment | added | Jan Weidner | I m not sure, whether canonical implies functorial. For example assign to an object of a category its automorphism group. This seems canonical to me, however it is not functorial. | |
| Mar 28, 2010 at 21:06 | comment | added | Dmitri Pavlov | @Konrad: For me natural = functorial = canonical. In particular, the natural isomorphism V → V** given by x ↦ (f ↦ -f(x)) is a canonical isomorphism for me, even though Reid Barton claims that it isn't. (I replaced 3 by -1 in his example to avoid problems with characteristic 3.) In my papers I avoid the term “canonical” and use more precise words such as “functorial” instead. | |
| Mar 28, 2010 at 19:43 | comment | added | François G. Dorais | @Dmitri: Yes, and your usage is correct, I apologize for the confusion. I'm leaving my comment there to since there is apparently a way to misunderstand what you wrote. (Though, after rereading carefully once more, I don't think what you wrote is ambiguous in any way. My guess is that accidentally skipping over a few words changes the meaning.) | |
| Mar 28, 2010 at 19:39 | comment | added | Konrad Waldorf | Your answer is the kind I was looking for, although I have the impression that you're talking about "natural" instead of "canonical". Or would you say that both words have the same meaning? Then you contradict Reid's answer above. | |
| Mar 28, 2010 at 19:17 | comment | added | Dmitri Pavlov | Qiaochu is right. I only said that “one cannot construct an isomorphism of functors id → * without using some form of the axiom of choice”, not that one cannot construct an isomorphism between a finite-dimensional vector space and its dual without using some form of the axiom of choice. | |
| Mar 28, 2010 at 19:16 | comment | added | François G. Dorais | @Qiaochu: You're right! I read too quickly. | |
| Mar 28, 2010 at 19:10 | comment | added | Qiaochu Yuan | To construct an isomorphism between the identity functor and the dual functor in the category of finite-dimensional vector spaces, isn't it necessary to choose, for each vector space, an explicit isomorphism to its dual? I think that's what Dmitri means. | |
| Mar 28, 2010 at 19:04 | comment | added | François G. Dorais | Your usage of "axiom of choice" is wrong -- the fact that a finite-dimensional vector space is isomorphic to its dual is provable without any use of the axiom of choice (even in topoi if you use the correct definition of finite). What you meant is probably "without making arbitrary choices" or "by making uniform choices." (The latter is how I usually understand the phrase "canonical choice.") | |
| Mar 28, 2010 at 18:52 | history | answered | Dmitri Pavlov | CC BY-SA 2.5 |