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1. • CommentRowNumber1.
   • CommentAuthorPeter Heinig
   • CommentTimeJun 14th 2017
   • (edited Jun 21st 2017)
   • PermaLink
   Author: Peter Heinig
   Format: MarkdownItex**What do you consider a usual technical term for the following kind of functor $P$?** Or rather, for the _pair_ of functors $(F,P)$ described by the list below? <sub> Can you point to parts of the literature where such a situation is studied more or less _in general_, for its own sake? (Although we do not expect there to be much to say about it in this generality, the literature _might_ hold surprises.) </sub> --------------------------------------------------------------------------------------------------------------------------------------- * $2$ is the _discrete_ category with two objects, * $\mathsf{C}$ is any (infinite) category, * $P\colon \mathsf{C}\rightarrow 2$ is any functor, * $F\colon\mathsf{C}\rightarrow\mathsf{C}$ is an endofunctor such that * * for each $i\in 2$ and each object $O$ of $\mathsf{C}$ with $P(O)=i$ we have $(P\circ F)(O)=1-i$. --------------------------------------------------------------------------------------------------------------------------------------- The latter property explains our (temporary) working terminology that $P$ is an _$F$-wise switchable property functor_: since $F(O)$ is an object of $\mathsf{C}$ again, we can switch between the two values of $P$ as often as we like, functorially, by applying $F$ over and over. (This implies obvious things, such as that $F$ cannot have any "fixed point", since the $O$ in the property after the black square is arbitrary). We also know that our endofunctor $F$ has the property that * * $F$ is not involutive, also not in weakened senses; the objects $O$, $F(O)$, $F(F(O))$ are quite "far" from each other, though thery are all _related_, because of: * * for each object $O$ of $\mathsf{C}$ there is at least one morphism $O\overset{f}{\rightarrow}F(O)$ in $\mathsf{C}$. (In various contexts there are other terms for such functors $F$, such as _inflationary operator_, _closure operator_ etc. This question is more about $P$ and the joint property it satisfies w.r.t. to $F$ than about this latter property that $F$ happens to have into the bargain.) Comments, more or less irrelevant to the question: * This question is partly motivated by ongoing graph-theoretical work with colleagues, applying methods from [https://arxiv.org/abs/1606.02926](https://arxiv.org/abs/1606.02926) to prove a non-axiomatizability-theorem for a large number of natural classes of _vertex-reconstructible_ graphs. We make essential use of an "operation" on the category of all countable forests, with isometric embeddings as the morphisms. It is essential that this "operation" actually is an endofunctor, and switches a property functor $P\colon\mathsf{C}\rightarrow2$ important to us. * It is easy to make up small artificial examples of such a situation. In a sense, the above situation occurs all over the place. (Though we are not aware of it being singled out as such, named and studied conceptually.) There also appear to be rather natural examples with $\mathsf{C}$ being some category of topological spaces and $F$ being a _suspension_-endofunctor, but using such examples in this question more than briefly touching upon them in this comment would be irrelevant and distracting. * The concrete category $\mathsf{C}$ that we are using is rather special from a categorical point of view, in particular, it is a [[left cancellative category]], but it is not at all finite. Spelling out these properties seems irrelevant to the question. * Needless to say, writing "functor $P\colon\mathsf{C}\rightarrow 2$" immediately implies that $P$ formalizes the usual notion of an _isomorphism-invariant property_: if there is an iso $O_0\overset{f}{\rightarrow}O_1$ in $\mathsf{C}$, then of course $P(O_0)\overset{P(f)}{\rightarrow}P(O_1)$ is an iso in $2$, so already its being a _morphism_ at all, together with $2$ being discrete, implies $P(O_0)=P(O_1)$. One could say that any functor into the _discrete_ category with two objects formalizes the concept of a _morphism_-invariant property: if there is a morphism between two objects, then they must either both have the property or both not have it.

   What do you consider a usual technical term for the following kind of functor $P$? Or rather, for the pair of functors $(F,P)$ described by the list below?

   _{Can you point to parts of the literature where such a situation is studied more or less in general, for its own sake? (Although we do not expect there to be much to say about it in this generality, the literature might hold surprises.)}

   ---

   • $2$ is the discrete category with two objects,

   • $\mathsf{C}$ is any (infinite) category,

   • $P\colon \mathsf{C}\rightarrow 2$ is any functor,

   • $F\colon\mathsf{C}\rightarrow\mathsf{C}$ is an endofunctor such that

   • • for each $i\in 2$ and each object $O$ of $\mathsf{C}$ with $P(O)=i$ we have $(P\circ F)(O)=1-i$.

   ---

   The latter property explains our (temporary) working terminology that $P$ is an $F$-wise switchable property functor: since $F(O)$ is an object of $\mathsf{C}$ again, we can switch between the two values of $P$ as often as we like, functorially, by applying $F$ over and over. (This implies obvious things, such as that $F$ cannot have any “fixed point”, since the $O$ in the property after the black square is arbitrary).

   We also know that our endofunctor $F$ has the property that

   • • $F$ is not involutive, also not in weakened senses; the objects $O$, $F(O)$, $F(F(O))$ are quite “far” from each other, though thery are all related, because of:
   • • for each object $O$ of $\mathsf{C}$ there is at least one morphism $O\overset{f}{\rightarrow}F(O)$ in $\mathsf{C}$.

   (In various contexts there are other terms for such functors $F$, such as inflationary operator, closure operator etc. This question is more about $P$ and the joint property it satisfies w.r.t. to $F$ than about this latter property that $F$ happens to have into the bargain.)

   Comments, more or less irrelevant to the question:

   • This question is partly motivated by ongoing graph-theoretical work with colleagues, applying methods from https://arxiv.org/abs/1606.02926 to prove a non-axiomatizability-theorem for a large number of natural classes of vertex-reconstructible graphs. We make essential use of an “operation” on the category of all countable forests, with isometric embeddings as the morphisms. It is essential that this “operation” actually is an endofunctor, and switches a property functor $P\colon\mathsf{C}\rightarrow2$ important to us.

   • It is easy to make up small artificial examples of such a situation. In a sense, the above situation occurs all over the place. (Though we are not aware of it being singled out as such, named and studied conceptually.) There also appear to be rather natural examples with $\mathsf{C}$ being some category of topological spaces and $F$ being a suspension-endofunctor, but using such examples in this question more than briefly touching upon them in this comment would be irrelevant and distracting.

   • The concrete category $\mathsf{C}$ that we are using is rather special from a categorical point of view, in particular, it is a left cancellative category, but it is not at all finite. Spelling out these properties seems irrelevant to the question.

   • Needless to say, writing “functor $P\colon\mathsf{C}\rightarrow 2$” immediately implies that $P$ formalizes the usual notion of an isomorphism-invariant property: if there is an iso $O_0\overset{f}{\rightarrow}O_1$ in $\mathsf{C}$, then of course $P(O_0)\overset{P(f)}{\rightarrow}P(O_1)$ is an iso in $2$, so already its being a morphism at all, together with $2$ being discrete, implies $P(O_0)=P(O_1)$. One could say that any functor into the discrete category with two objects formalizes the concept of a morphism-invariant property: if there is a morphism between two objects, then they must either both have the property or both not have it.

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 14th 2017
   • (edited Jun 14th 2017)
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   Author: Todd_Trimble
   Format: MarkdownItexPutting $Ob(2) = \{0, 1\}$, it seems to be tantamount to a pair of functors $G: C_0 \to C_1, H: C_1 \to C_0$. Given such a $(P, F)$, let $C_0$ be the fiber over $0$ and $C_1$ the fiber over $1$, etc. In the other direction, given the pair $G, H$, let $C = C_0 + C_1$, and define $F: C \to C$ by $G$ on $C_0$ and $H$ on $C_1$, etc. Of course you know this already. I can't think of a name from the literature, but a whimsical name might be "pisces-functor". It amounts to a quiver map $(a \leftrightarrows b) \to Cat$ where $a, b$ are the "two fish".

   Putting $Ob(2) = \{0, 1\}$, it seems to be tantamount to a pair of functors $G: C_0 \to C_1, H: C_1 \to C_0$. Given such a $(P, F)$, let $C_0$ be the fiber over $0$ and $C_1$ the fiber over $1$, etc. In the other direction, given the pair $G, H$, let $C = C_0 + C_1$, and define $F: C \to C$ by $G$ on $C_0$ and $H$ on $C_1$, etc. Of course you know this already.

   I can’t think of a name from the literature, but a whimsical name might be “pisces-functor”. It amounts to a quiver map $(a \leftrightarrows b) \to Cat$ where $a, b$ are the “two fish”.