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Stream: deprecated: id my structure

Topic: almost a full subcategory?

Joshua Meyers (Jan 04 2022 at 05:49):

Let $M$ be the category of metric spaces and Lipschitz maps. Let $P$ be the category whose objects are spaces $X$ equipped with a function $d:X\times X\to\mathbb{R}$ (no axioms) and whose morphisms from $(X,d)$ to $(X',d')$ are ordered pairs $(\alpha,f:X\to Y)$ where $f$ satisfies $\forall x,y\in X\ldotp d'(f(x),f(y))<\alpha d(x,y)$. I'd like to say that $M$ embeds as a full subcategory of $P$, but that is not the case. So what is going on here?

I have thought about this a bit: considering the canonical functor $F:M\to P$ sending $(X,d)$ to $(X,d)$ and $f$ to $(\alpha,f)$, where $\alpha$ is the smallest value such that $f$ is $\alpha$-Lipschitz, let $M'$ be the full subcategory of $P$ spanned by the image of $F$. Then the crucial fact is that the functor $F:M\to M'$ has a "pseudo-retraction", i.e. a functor $P:M'\to M$ such that $P\circ F\cong {\sf id}_M$. Oddly, I can't find any literature on such "pseudo-retractions". Anyone know of anything?

Joshua Meyers (Jan 04 2022 at 06:04):

A more trivial example of this same thing is the functor $C\to C\times 1\to C\times D\to C\times D+E$, for categories $C$, $D$, and $E$.

Todd Trimble (Jan 04 2022 at 14:42):

Joshua Meyers said:

Let $M$ be the category of metric spaces and Lipschitz maps. Let $P$ be the category whose objects are spaces $X$ equipped with a function $d:X\times X\to\mathbb{R}$ (no axioms) and whose morphisms from $(X,d)$ to $(X',d')$ are ordered pairs $(\alpha,f:X\to Y)$ where $f$ satisfies $\forall x,y\in X\ldotp d'(f(x),f(y))<\alpha d(x,y)$. I'd like to say that $M$ embeds as a full subcategory of $P$, but that is not the case. So what is going on here?

I have thought about this a bit: considering the canonical functor $F:M\to P$ sending $(X,d)$ to $(X,d)$ and $f$ to $(\alpha,f)$, where $\alpha$ is the smallest value such that $f$ is $\alpha$-Lipschitz, let $M'$ be the full subcategory of $P$ spanned by the image of $F$. Then the crucial fact is that the functor $F:M\to M'$ has a "pseudo-retraction", i.e. a functor $P:M'\to M$ such that $P\circ F\cong {\sf id}_M$. Oddly, I can't find any literature on such "pseudo-retractions". Anyone know of anything?

One thing that's not clear to me is that that smallest $\alpha$ exists.

Kenji Maillard (Jan 04 2022 at 14:59):

Shouldn't it be $\leq$ rather than $<$ in the condition $\forall x,y \in X. d'(f(x), f(y)) < \alpha d(x,y)$ ? Otherwise multiplication by two $-\times 2 : [0;1] \to [0;2]$ does not satisfy the condition for $\alpha = 2$.
And even when a smallest $\alpha$ exists, there is no reason that it would be functorial: if $f(x) = 2x : \{0,1\} \to \{0,1,2\}$ and $g(y) = \min(2, 2y) : \{0,1,2\}\to \{0,2\}$ then $P(f) = (2, f), P(g) = (2, g)$ but $P(g \circ f) = (2, g\circ f) \neq (2*2, g\circ f) = P(g) \circ P(f)$ (assuming composition in $M$ multiplies the coefficients)

Joshua Meyers (Jan 04 2022 at 15:52):

good points people, yeah it should be $\leq$ rather than $<$; then a smallest $\alpha$ always exists and is given by $\sup_{x\neq y}\frac{d'(f(x),f(y))}{d(x,y)}$. But yeah, "smallest $\alpha$" is not preserved under composition, which causes problems.

Joshua Meyers (Jan 04 2022 at 15:58):

So then what is the relationship between $P$ and $M$?

Joshua Meyers (Jan 04 2022 at 16:02):

Oh wow I can't believe it I think it's a cofunctor, as in @David Spivak's poly book! The mapping $F:M\to P$ goes forward on objects and then maps morphisms backwards $P(FX,FX')\to M(X,X')$ by $(\alpha,f)\mapsto f$.

Joshua Meyers (Jan 04 2022 at 16:03):

What's more it's a "full cofunctor", which I define as a cofunctor where the backwards morphism maps are surjective

Joshua Meyers (Jan 04 2022 at 16:09):

full functors $\supseteq$ fully faithful functors $=$ fully faithful cofunctors $\subseteq$ full cofunctors

Joshua Meyers (Jan 04 2022 at 16:10):

The middle equality is because when $M(X,X')\cong P(FX,FX')$, the isomorphism can be thought of in either direction.

Joshua Meyers (Jan 04 2022 at 16:10):

Thus full cofunctors strictly generalize full embeddings

Joshua Meyers (Jan 04 2022 at 16:24):

Whoops I'm wrong! To be a cofunctor it would have to send any morphism $FX\to Y$ in $P$ back to a morphism $X\to X'$ for some $X'$ such that $FX'=Y$. But this is impossible as $F$ is not surjective on objects.

Joshua Meyers (Jan 04 2022 at 16:26):

OK I now think it's a span: $M\xleftarrow{I} M'\xrightarrow{F} P$, where $I$ is full and essentially surjective and $F$ is full and faithful.

Joshua Meyers (Jan 04 2022 at 16:27):

$M'$ here is the full subcategory of $P$ spanned by metric spaces.

Joshua Meyers (Jan 04 2022 at 16:42):

But this still isn't all the structure that's there...