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Stream: learning: questions

Topic: lifting properties

Matteo Capucci (he/him) (Aug 25 2022 at 15:06):

Hi folks,
I asked this question on Twitter, let me echo it here: https://twitter.com/mattecapu/status/1562816006204321793?s=20&t=HAdCYkJQzKBCT_ukhej4ag

can anyone motivate why lifting properties are stated for squares instead of cospans? i.e. why i and f are there in this def? https://twitter.com/mattecapu/status/1562816006204321793/photo/1

- Matteo Capucci (@mattecapu)

Matteo Capucci (he/him) (Aug 25 2022 at 15:08):

To clarify, I'm asking why lifting properties are not stated as 'for every $g$ and $p$ of (some nice class of morphisms) there is a lift $q$ such that (bunch of properties)'

Matteo Capucci (he/him) (Aug 25 2022 at 15:09):

image.png

Tobias Schmude (Aug 25 2022 at 15:18):

I'd like to know that as well! The only "motivation" I know of is that it subsumes lifting and extension properties.

Reid Barton (Aug 25 2022 at 15:19):

This was already basically answered on Twitter, but the simplest example from topology would be that we have a map $p : X \to Y$, and we want to lift a path $\gamma$ in $Y$ to a path in $X$ that starts at a specified lift of the initial point of $\gamma$. Diagrammatically, that is a condition involving a square with $\{0\} \to [0, 1]$ and $p : X \to Y$ as its two sides.

Reid Barton (Aug 25 2022 at 15:35):

Note that the condition is about $i$ and $p$, and quantifies over all $f$ and $g$. If $A$ is the initial object, so that the diagram reduces to the one you drew, it would still be a condition about $p$ that quantifies over all $g$, not a condition about all $p$ and $g$.

Mike Shulman (Aug 25 2022 at 18:56):

On the other hand, lifting in a square is the same as lifting in a cospan in a coslice category...

Morgan Rogers (he/him) (Aug 25 2022 at 20:07):

Might be worth observing that with different quantification on the cospan you can instead define [[projective object]] s.

John Baez (Aug 25 2022 at 20:57):

Oh, I completely misunderstood your question on Twitter, Matteo. I thought you were wanting to do something really weird.

John Baez (Aug 25 2022 at 20:58):

In fact, the kind of lifting property you're talking about is very commonly used and important. For example it shows up in the definition of [[projective module]].

John Baez (Aug 25 2022 at 21:01):

However as Barton pointed out, in topology (and elsewhere) we also often want to look at a combination of lifting and extension. His example is the fundamental one: we are trying to lift a path in $Y$ to a path in $X$, but we have already chosen how to lift the starting point of that path, so we are extending that lift to a lift of the whole path.

John Baez (Aug 25 2022 at 21:02):

This example should remind you of the concept of Grothendieck (op)fibration.

Matteo Capucci (he/him) (Aug 26 2022 at 07:44):

Thanks everyone for the replies :) I now see the point behind a square, as a lift with an extension condition (or an extension with a lifting condition)

Matteo Capucci (he/him) (Aug 26 2022 at 07:48):

Yesterday I was playing with actions and spans and I wondered what the action of a monad in span(C) on another endomorphism looks like
Monads in span(C) are categories, so this almost recovers actions of categories, but there's an extra condition. Indeed what you get is the 'action of a category on a graph', i.e. you get a composition operation that takes an arrow of a graph (internal to C) and extends it with an arrow of the acting category: image.png

Matteo Capucci (he/him) (Aug 26 2022 at 07:48):

(It is crucial that both the graph and the category share the same set of vertices/objects)

Matteo Capucci (he/him) (Aug 26 2022 at 07:49):

Call the acting category $\cal M$. The thing above is an action of $\cal M$ on a graph $G : C \leftarrow E \to C$

Matteo Capucci (he/him) (Aug 26 2022 at 07:50):

If we swap the legs of $\cal M$ we get $\cal M^{op}$. Now a (right) action of $\cal M^{op}$ on $G$ is an operation such as

Matteo Capucci (he/him) (Aug 26 2022 at 07:50):

image.png

Matteo Capucci (he/him) (Aug 26 2022 at 07:51):

and we can read $f \bullet m$ as 'the lift of $f$ along $m$'

Matteo Capucci (he/him) (Aug 26 2022 at 07:52):

The laws of action tell us that $f \bullet (m ; n) = (f \bullet m) ; n$ and $f \bullet 1 = f$, which seem reasonable conditions for a lifting operation

Matteo Capucci (he/him) (Aug 26 2022 at 07:52):

Dually, a left action of $\cal M^{op}$ gives an extension operation:

Matteo Capucci (he/him) (Aug 26 2022 at 07:52):

image.png

Matteo Capucci (he/him) (Aug 26 2022 at 07:52):

with similar properties

Matteo Capucci (he/him) (Aug 26 2022 at 07:54):

Now I wonder if 'lifting properties' expressed as squares can be expressed in this way... I tried figuring out what a bimodule is but it falls short of being the right thing

Mike Shulman (Aug 26 2022 at 15:13):

A bimodule (endo- or not) between monads M, N in $\rm Span(Set)$ is a [[profunctor]] $M^{op}\times N\to \rm Set$. A bimodule in ${\rm Span}(C)$ is an [[internal profunctor]] in $C$. A one-sided module is the same as a bimodule with a trivial monad on one side.

Matteo Capucci (he/him) (Aug 28 2022 at 14:19):

Aaah sure :O that's cool