#postcomposition

[bubbloquacious]

Hydrogen bomb vs. coughing baby: graphs and the Yoneda embedding

So we all love applying heavy duty theorems to prove easy results, right? One that caught my attention recently is a cute abstract way of defining graphs (specifically, directed multigraphs a.k.a. quivers). A graph G consists of the following data: a set G(V) of vertices, a set G(A) of arrows, and two functions G(s),G(t): G(A) -> G(V) which pick out the source and target vertex of an arrow. The notation I've used here is purposefully suggestive: the data of a graph is exactly the same as the data of a functor to the category of sets (call it Set) from the category that has two objects, and two parallel morphisms from one object to the other. We can represent this category diagrammatically as ∗⇉∗, but I am just going to call it Q.

The first object of Q we will call V, and the other we will call A. There will be two non-identity morphisms in Q, which we call s,t: V -> A. Note that s and t go from V to A, whereas G(s) and G(t) go from G(A) to G(V). We will define a graph to be a contravariant functor from Q to Set. We can encode this as a standard, covariant functor of type Q^op -> Set, where Q^op is the opposite category of Q. The reason to do this is that a graph is now exactly a presheaf on Q. Note that Q is isomorphic to its opposite category, so this change of perspective leaves the idea of a graph the same.

On a given small category C, the collection of all presheaves (which is in fact a proper class) has a natural structure as a category; the morphisms between two presheaves are the natural transformations between them. We call this category C^hat. In the case of C = Q, we can write down the data of such a natural transformations pretty easily. For two graphs G₁, G₂ in Q^hat, a morphism φ between them consists of a function φ_V: G₁(V) -> G₂(V) and a function φ_A: G₁(A) -> G₂(A). These transformations need to be natural, so because Q has two non-identity morphisms we require that two specific naturality squares commute. This gives us the equations G₂(s) ∘ φ_A = φ_V ∘ G₁(s) and G₂(t) ∘ φ_A = φ_V ∘ G₁(t). In other words, if you have an arrow in G₁ and φ_A maps it onto an arrow in G₂ and then you take the source/target of that arrow, it's the same as first taking the source/target in G₁ and then having φ_V map that onto a vertex of G₂. More explicitly, if v and v' are vertices in G₁(V) and a is an arrow from v to v', then φ_A(a) is an arrow from φ_V(v) to φ_V(v'). This is exactly what we want a graph homomorphism to be.

So Q^hat is the category of graphs and graph homomorphisms. This is where the Yoneda lemma enters the stage. If C is any (locally small) category, then an object C of C defines a presheaf on C in the following way. This functor (call it h_C for now) maps an object X of C onto the set of morphisms Hom(X,C) and a morphism f: X -> Y onto the function Hom(Y,C) -> Hom(X,C) given by precomposition with f. That is, for g ∈ Hom(Y,C) we have that the function h_C(f) maps g onto g ∘ f. This is indeed a contravariant functor from C to Set. Any presheaf that's naturally isomorphic to such a presheaf is called representable, and C is one of its representing objects.

So, if C is small, we have a function that maps objects of C onto objects of C^hat. Can we turn this into a functor C -> C^hat? This is pretty easy actually. For a given morphism f: C -> C' we need to find a natural transformation h_C -> h_C'. I.e., for every object X we need a set function ψ_X: Hom(X,C) -> Hom(X,C') (this is the X-component of the natural transformation) such that, again, various naturality squares commute. I won't beat around the bush too much and just say that this map is given by postcomposition with f. You can do the rest of the verification yourself.

For any small category C we have constructed a (covariant) functor C -> C^hat. A consequence of the Yoneda lemma is that this functor is full and faithful (so we can interpret C as a full subcategory of C^hat). Call it the Yoneda embedding, and denote it よ (the hiragana for 'yo'). Another fact, which Wikipedia calls the density theorem, is that any presheaf on C is, in a canonical way, a colimit (which you can think of as an abstract version of 'quotient of a disjoint union') of representable presheaves. Now we have enough theory to have it tell us something about graphs that we already knew.

Our small category Q has two objects: V and A. They give us two presheaves on Q, a.k.a. graphs, namely よ(V) and よ(A). What are these graphs? Let's calculate. The functor よ(V) maps the object V onto the one point set Hom(V,V) (which contains only id_V) and it maps A onto the empty set Hom(A,V). This already tells us (without calculating the action of よ(V) on s and t) that the graph よ(V) is the graph that consists of a single vertex and no arrows. The functor よ(A) maps V onto the two point set Hom(V,A) and A onto the one point set Hom(A,A). Two vertices (s and t), one arrow (id_A). What does よ(A) do with the Q-morphisms s and t? It should map them onto the functions Hom(A,A) -> Hom(V,A) that map a morphism f onto f ∘ s and f ∘ t, respectively. Because Hom(A,A) contains only id_A, these are the functions that map it onto s and t in Hom(V,A), respectively. So the one arrow in よ(A)(A) has s in よ(A)(V) as its source and t as its target. We conclude that よ(A) is the graph with two vertices and one arrow from one to the other.

We have found the representable presheaves on Q. By the density theorem, any graph is a colimit of よ(V) and よ(A) in a canonical way. Put another way: any graph consists of vertices and arrows between them. I'm sure you'll agree that this was worth the effort.

[notarealwelder]

Say we have some continuous functions f, f' : X → Y and g, g' : Y → Z, and homotopies F : X×I → Y : f ≃ f' and G : Y×I → Z : g ≃ g'.

We might guess that f;g, f';g' : X → Z are homotopic also, and we would be right! Homotopic how exactly, though?

As Is Known, if we pre/postcompose a function to a homotopy, we get a homotopy of pre/postcompositions. And of course homotopies are transitive. So we can do either of:

f;G : f;g ≃ f;g' and F;g' : f;g' ≃ f';g', so f;G `trans` F;g' : f;g ≃ f';g'

F;g : f;g ≃ f';g and f';G : f';g ≃ f';g', so F;g `trans` f';G : f;g ≃ f';g'

In the first we kind of "do" image-of-G first and image-of-F second, and in the second, the other way around.

But wait, it's worse! We can do them simultaneously! Whyever not!

let H x t = G (F x t) t. then H x 0 = G (F x 0) 0 = g (f x), and H x 1 = G (F x 1) 1 = g' (f' x), so H : f;g ≃ f';g'.

So, questions:

what the fuck?

and which one of these is the....right one to use?

[vinyldinosaur]

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———————————————— steps towards showing un- begged, borrowed, or stolen art; through reused things. ——————————————— #instrumentassistedsofttissuemanipulation #quickmades #magicbanality #demo #jazz #yamahadjx #consumertechnology #bedroomproducer #sitespecific #postcomposition #haiku #poemlet #radicallive #marketingstrategy (at Wedding, Berlin) https://www.instagram.com/p/B1ZUVjzoZdb/?igshid=zi2e58zcw4sv

[vinyldinosaur]

I

Will

Try

To

Continue

My

Musicianship blog

Kill daddy [canonbeing]

I went to a performance art table One being wanted to kill Beuys; which #canonicaling of your #particular Discipline Do you want to knock off?

[as-if-and-only-if]

So you know how you can talk about pre- or postcomposing with a function as an operation in itself, usually denoted as f* such that f*(g) = g ∘ f for precomposition and f∗ such that f∗(g) = f ∘ g for postcomposition? And then you need a name for f* and f∗, and people call those the “pullback” and “pushforward” sometimes respectively, but it’s kind of confusing because those are also names for more general constructions and it’s not clear whether “the pullback of f” is a thing that pulls back along f (common intended meaning), or is what you get by pulling f back along something else (which means the opposite)? And calling them “precomposition with f” and “postcomposition with f” is long and tiresome?

so here’s my proposed nomenclature, to take advantage of English’s syntax: f* is pronounced “f-pre” and f∗ is pronounced “f-post”

Then f*(g) is pronounced “f-pre g” (oh my god that looks like “fpreg” but just. say it out loud instead pls) and means what it sounds like: “f, pre-g”. See how “g” is now in position to be the complement of the preposition “pre”, and the whole phrase “pre-g” can be an adjunct of “f”? Because f*(g) means that f is pre-g? Same with f∗(g) as “f-post g” = “f, post-g”, and f is post-g. So there ya go!

[mathematicianadda]

Cohomology ring of $H^*(\mathbb{R}P^{\infty} \times \mathbb{R}P^{\infty}; \mathbb{Z})$

This is course work

Prove that the cohomology ring $H^*(\mathbb{R}P^{\infty} \times \mathbb{R}P^{\infty}; \mathbb{Z})$ is isomorphic to the subring of $\mathbb{Z}[x, y]/(2x, 2y)$ generated by $x^2, y^2, xy^2 + x^2y$ with $x$ and $y$ in degree $1$.

Hint: Use the Bockstein cohomology to show that every element of $\tilde{H}^*(\mathbb{R}P^\infty \times \mathbb{R}P^\infty; \mathbb{Z})$ is of order $2$.

I think the Bockstein cohomology is trivial, but have trouble showing it.

We have a projection $\pi: \mathbb{R}P^\infty \times \mathbb{R}P^\infty \to \mathbb{R}P^\infty$ with a section $x \mapsto (x, y_0)$ where $y_0$ is the zero-cell of $\mathbb{R}P^\infty$. So the induced homomorphism is split mono, hence injective. Using naturality to obtain a diagram

$\require{AMScd}$ \begin{CD} H^n(\mathbb{R}P^\infty; \mathbb{Z}/2) @>{\beta'}>> H^{n+1}(\mathbb{R}P^\infty; \mathbb{Z}/2)\\ @V{\pi^*}VV @V{\pi^*}VV\\ H^{n}(\mathbb{R}P^\infty \times \mathbb{R}P^\infty; \mathbb{Z}/2) @>{\beta}>> H^{n+1}(\mathbb{R}P^\infty \times \mathbb{R}P^\infty; \mathbb{Z}/2) \end{CD}

where $\beta$ and $\beta'$ are Bocksteins. It is known that $\beta'$ is an isomorphism for $n$ odd, as $\pi^*$ is injective, we see that the kernel of $\beta$ is contained in the image of $\pi^*$ (the left one). For $n$ even, it is known that $\beta'$ is $0$. Taking the naturality square for $i^*$ we see that $i^* \circ \beta$ is the zero map. So the image of $\beta$ is contained in the kernel of $i^*$ (the left one). Here I am stuck.

Suppose that the Bockstein cohomology is trivial. We can write any abelian group/$\mathbb{Z}$-module as a direct sum of copies of $\mathbb{Z}$ and $\mathbb{Z}/p^k$ for $p$ prime, and know that postcomposition with $\mathbb{Z} \to \mathbb{Z}/p$ sends summands $\mathbb{Z}$ in $H^n(X; \mathbb{Z}$ to a summand $\mathbb{Z}/p$ in $BH_p^n(X)$, a summand $\mathbb{Z}/p^k$ in $H^n(X, \mathbb{Z})$ to a summand $\mathbb{Z}/p$ in $BH_p^n(X)$ for $k > 1$. If the Bokstein cohomology is trivial, then $H^*(\mathbb{R}P^\infty \times \mathbb{R}P^\infty; \mathbb{Z})$ must consist out of copies of $\mathbb{Z}/2$, which proves the statement of the hint.

We have calculated before that $H^*(\mathbb{R}P^\infty; \mathbb{Z}) \cong \mathbb{Z}[x]/(2x)$, so the projections induce injections $\mathbb{Z}[x]/(2x), \mathbb{Z}[y]/(2y)$ into $H^*(\mathbb{R}P^\infty \times \mathbb{R}P^\infty; \mathbb{Z})$. I don't know how to continue.

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/3ajLJac

[bubbloquacious]

I'm struggling a little bit to find the right 'data type' for a scissors morphism (of which a scissors congruence would be an isomorphism) in an arbitrary category C. Because the arrows of C aren't necessarily functions, you can't have a scissors morphism just be a more lax notion of function. The heavy handed way to do it would be to have a scissors morphism be an indexed family of C-morphisms, defined on a covering family of its domain object, satisfying certain coherence properties (this is how Zakharevich does it, or something like it). But that feels bad! It's too extrinsic, and I think you run the risk of having multiple morphisms for what 'should' be the same action.

One thing I tried is to have it be a specific 'sieve correspondence'. A sieve is kind of like a subobject, except there's way more of them. Say your morphisms are inclusions of squares into squares in the plane. There's no subobject in this category corresponding to a disk inside the square, but there is the sieve of all squares contained within that disk. Specifically, you can identify sieves on an object with the subfunctors of the representable presheaf of that object. So you can say that the data type of a scissors morphism is a (possibly union-preserving, to make it function-like) action from the sieves on the domain to the sieves on the codomain!

There's an issue here, though. Including the C-morphisms as trivial scissors morphisms (for the covering family consisting of just the identity arrow) in this way is not generally faithful. If you have the category of polygons and isometric inclusions then it is faithful, because an isometry on a figure with interior is uniquely determined by its action on the subfigures (so certainly by the action on the sieves!). But for the cyclic group of order 2 considered as a one-object category you run into issues. The presheaves on this category are the C₂-sets and the unique representable presheaf is the two point set with the obvious action. This object has two automorphisms, but both have the same action on the subobjects, because neither of the points are invariant subsets on their own. So reducing to sieve actions is not faithful!

So famously one of the consequences of the Yoneda lemma is that the action on representable presheaves (i.e. homset functors with fixed codomain) of postcomposition by some arrow is a faithful representation of the category; if postcomposing by f, f': A -> B gives the same function Hom(X,A) -> Hom(X,B) for all objects X, then f = f'. The problem is that it is also full; all such natural transformations are given by some postcomposition, specifically postcomposition by the image of id ∈ Hom(A,A) (and that's the proof of the Yoneda lemma, by the way, that's why people say it's so simple).

So maybe just straight up transformations (sans naturality) between representable presheaves? That is, just a set function Hom(X,A) -> Hom(X,B) for every object X? The problem is that you then need to decide where id_A gets sent to. And the 'image' of a scissors morphism that you do want to exist might not be a subobject. Like, if your category only contains connected polygons, you still want the scissors morphisms to contain actions that disconnect the object. That's what scissors do! So again mapping it onto a sieve is perfect, since it contains exactly those phantom subobjects that you need to exist. You just need enough sieves to differentiate the C-arrows. But how to get them?

[vinyldinosaur]

Visual Missive Collecting

    Bringing together threads so it's out-of-the-box for you, it's out-of-the box and straight onto your screen.

A collection of my first person video pieces, collecting from my wanderings around a certain part of a certain city.

[vinyldinosaur]

Ephemeral documentation - proof of artistic effort #dailygeneralreport #bust #mozart #postcomposition #videoart #workinprogress https://www.instagram.com/p/B0lSclGIyst/?igshid=15rynme1hu5ir

[vinyldinosaur]

#portfolio #career #icecube #consortium #selfieportrait #badge #door #postcomposition (at Wedding, Berlin) https://www.instagram.com/p/B0SxsxjIafg/?igshid=42kjd8qmmxu9

[vinyldinosaur]

Vinyl Dinosaur #ScienceCommunication #division - making #nofuture since sometime in the #70s #timefowls #dinosaur #fowl #anthropocene #designideas #albumcover #evolution #postcomposition chicken taken off flickr: https://flic.kr/p/7swLjj https://www.instagram.com/p/Bz2Sq_YITEG/?igshid=2nu60kobqt1x

[vinyldinosaur]

lots happens between fat times of making large sounds; coffee and pastries meetings fundraising governmental red-tape and branding meet-and-greets preparations for the daily general meeting with the share-holders the consortium feels that annual reporting is quite démodé today there have been great strides in the production of electric sound #precomposition #postcomposition #haiku #instapoetry #poemlet #consortium #meetings #paperwork #antibiggovernment #librarymusic #pastries #coffee #electronica (at Wedding, Berlin) https://www.instagram.com/p/BzvUDhHIlpE/?igshid=t3ar1dt02ncd

[vinyldinosaur]

the consortium is brainstorming ways to bring clients to the fold #consumertechnology #brand #consortium #vinyldinosaur #vindindustry #affectivespaces #narration #customerengagement #amateurprofessional #calligraphy #asocialmedia #postcomposition (at Schaubühne Lindenfels) https://www.instagram.com/p/BzYaStjoleS/?igshid=1h2r6ea5ebnek

[vinyldinosaur]

make it #simple but #significant 

https://www.instagram.com/p/BxZ_k2Tgqhs/

[mathematicianadda]

Exponents in a slice category of presheaves

Let $\mathcal{C}$ be a category and denote $\hat{\mathcal{C}}$ for the category of presheaves on $\mathcal{C}$. For $k: K \to F$, denote $k^*$ for the pullback functor $\hat{\mathcal{C}}/F \to \hat{\mathcal{C}}/K$. Denote $\Sigma_k$ for the postcomposition by $k$ as a functor $\hat{\mathcal{C}}/K \to \hat{\mathcal{C}}/F$ and note that $\Sigma_k \dashv k^*$.

I want to find exponents in $\hat{\mathcal{C}}/F$. For some $g: G \to F$, $h: H \to F$, $k: K \to F$, we must have that

$$\begin{align} \hat{\mathcal{C}}/F(k, h^g) &\simeq \hat{\mathcal{C}}/F(k \times g, h)\\ & \simeq \hat{\mathcal{C}}/F(\Sigma_k \circ k^*(g), h) \\ &\simeq \hat{\mathcal{C}}/K(k^*(g), k^*(h)). \end{align}$$

For $k$ the identity on $F$, this says that the sections of $h^g$ are in natural bijective correspondence with $\mathcal{C}/F(g, h)$ which is neat I suppose, but does not bring us a whole lot further.

I want to take $k$ a morphism between the Yoneda embedding $y_C$ and $F$ (aka an element of $FC$) for a bunch of $C$'s and try to use the Yoneda lemma and the fact that morphisms in $\hat{\mathcal{C}}/F$ are morphisms in $\hat{\mathcal{C}}$ in particular, but have not managed it, and for $C$'s such that $FC = \emptyset$, they do not even exist.

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/2v1k2DQ