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Stream: learning: questions

Topic: Commuting Square functions.

Keith Elliott Peterson (Nov 07 2024 at 21:55):

I tried to find the expontential object for the topos $Set^{\rightarrow}$.

The solution I found was,

$$f^g:: \sum_{h\in{\text{cod}(f)}^{\text{cod}(g)}} \prod_{i\in\text{cod}(g)} f^{-1}(i)^{g^{-1}(I)} \rightarrow {\text{cod}(f)}^{\text{cod}(g)}.$$

It certainly looks the part.

I suppose this means that just as there are function sets (in $Set$) there is also a commuting square sets (in $Set$). In fact, commuting n-hypercube sets as well. Neat, but nothing surprising.

My question is, how does this extend to exponential objects in categories of presheaves for other ordinals? Specifically, $Set^{\omega}$ and $Set^{\omega^{op}}$ (and I guess of other limit ordinals as well).

Kevin Carlson (Nov 08 2024 at 00:47):

There's a general formula for exponentials in presheaf categories: $X^Y(c)=\mathrm{Set}^C(\hat c,X^Y)=\mathrm{Set}^C(Y\times \hat c,X)$, where $X,Y:C\to\mathrm{Set}, c$ is an object of $C,$ and $\hat c$ is the representable functor. For ordinals, the representables are either downward- or upward-closed chains and so $Y\times \hat c$ just erases one "end" or the other of $Y$; maybe you'd like to see whether you can find a conclusion to the computation that satisfies your understanding.

Morgan Rogers (he/him) (Nov 08 2024 at 14:19):

What a coincidence, I was discussing exponentials in $\mathrm{Set}^{\omega^{op}}$ with Louis Lemonnier yesterday.

Matteo Capucci (he/him) (Nov 08 2024 at 14:19):

it's that time of the year

Keith Elliott Peterson (Nov 09 2024 at 21:26):

Kevin Carlson said:

There's a general formula for exponentials in presheaf categories: $X^Y(c)=\mathrm{Set}^C(\hat c,X^Y)=\mathrm{Set}^C(Y\times \hat c,X)$, where $X,Y:C\to\mathrm{Set}, c$ is an object of $C,$ and $\hat c$ is the representable functor. For ordinals, the representables are either downward- or upward-closed chains and so $Y\times \hat c$ just erases one "end" or the other of $Y$; maybe you'd like to see whether you can find a conclusion to the computation that satisfies your understanding.

I know, but turning $[Y\times\hat c, X]$ is the challenge. :P

My assumption is that an exp. object in something like $Set^{\overrightarrow{n}}$ would be a generalization of an exponential object in $Set^\rightarrow \cong Set^{\overrightarrow{2}}$.

I'm not exactly sure what you mean by "erasing one end." Is not a limit ordinal precisely an ordinal (reversed or not) that has no elements "erased?"

Matteo Capucci (he/him) said:

it's that time of the year

Topos theory has a holiday this time of year?

Todd Trimble (Nov 09 2024 at 23:22):

Topos theory has a holiday this time of year?

The way I read the joke is that $\mathrm{Set}^{\omega^{op}}$ is equivalently the category of trees. :-)

Ryuya Hora (Nov 10 2024 at 01:48):

Keith Elliott Peterson said:

I tried to find the expontential object for the topos $Set^{\rightarrow}$.

The solution I found was,

$$f^g:: \sum_{h\in{\text{cod}(f)}^{\text{cod}(g)}} \prod_{i\in\text{cod}(g)} f^{-1}(i)^{g^{-1}(I)} \rightarrow {\text{cod}(f)}^{\text{cod}(g)}.$$

I guess you meant

$$f^g:: \sum_{h\in{\text{cod}(f)}^{\text{cod}(g)}} \prod_{i\in\text{cod}(g)} f^{-1}(h(i))^{g^{-1}(i)} \rightarrow {\text{cod}(f)}^{\text{cod}(g)}.$$

Ryuya Hora (Nov 10 2024 at 02:22):

I agree that the general formula of exponentials in presheaf topos is practical to some extent to this particular example. Here, I would like to mention something a little more non-general phenomenon here.

Observation: An object of the topos $\mathrm{Set}^{\to}$ can be regarded as an indexed family (= formal coproduct) of sets $\{X_i\}_{i\in I}$. From this point of view, we have

$\{X_i\}_{i\in I}^{\{Y_j\}_{j\in J}} = \prod_{j\in J} \{X_i^{Y_j}\}_{i\in I}$.

We can generalize this observation. For a small category $C$, we can consider a new category $C^\triangleleft$ equipped with a new formal initial object. Finite ordinals are recursively obtained by $n+1 = n^{\triangleleft}$.
My claim is: $\mathrm{PSh}(C^\triangleleft)$ is the category of families of objects of $\mathrm{PSh}(C)$, and the exponentials in $\mathrm{PSh}(C^\triangleleft)$ is calculated by the same formula above with the exponentials in $\mathrm{PSh}(C)$.

(Here, we use not only the "exponential laws", but also the fact that the topos $\mathrm{PSh}(C^{\triangleleft})$ is stably locally connected. This ensures that the exponential functor ${-}^X\colon \mathcal{E}\to \mathcal{E}$ for a connected object $X$ preserves coproducts.)

This allows us to calculate exponentials in $\mathrm{PSh}(n)$ inductively.
Even for $\omega$, we can calculate exponentials with this formula from lower index $n\in \omega$. Exponential at index $n\in \omega$ does not depend on data of bigger indices $m>n$ since the forgetful functor $\mathrm{PSh}(\omega) \to \mathrm{PSh}(n)$ preserves exponentials (since this is a sheafification functor for an open subtopos).

This method get stuck for $\omega$ since $\omega^{\triangleleft}= \omega$.

Ryuya Hora (Nov 10 2024 at 02:26):

I think the reason why there appear the commutative diagrams is:
$\mathrm{dom}(f^g) \cong \mathrm{Set}^{\to}(1, f^g) \cong \mathrm{Set}^{\to}(g, f)\cong \text{Commutative squares from }g \text{ to }f$.

Keith Elliott Peterson (Nov 11 2024 at 22:04):

Todd Trimble said:

The way I read the joke is that $\mathrm{Set}^{\omega^{op}}$ is equivalently the category of trees. :-)

Ah! I usually think of $\mathrm{Set}^{\omega^{op}}$ as the category of forests, so it went over my head (even though of course $\mathrm{Set}^{\omega^{op}} \cong \mathrm{Set}^{(1+\omega)^{op}}$). Heh, clever.

So is a Christmas tree an object of $\mathrm{Set}^{\omega^{op}}/\mathbf{xmas}$? :laughing:

Ryuya Hora said:

I guess you meant

$$f^g:: \sum_{h\in{\text{cod}(f)}^{\text{cod}(g)}} \prod_{i\in\text{cod}(g)} f^{-1}(h(i))^{g^{-1}(i)} \rightarrow {\text{cod}(f)}^{\text{cod}(g)}.$$

Oh, yes, good catch. Correct, we want to pullback $h(i)$ not just $i$ along $f$.

Ryuya Hora said:

I agree that the general formula of exponentials in presheaf topos is practical to some extent to this particular example. Here, I would like to mention something a little more non-general phenomenon here.

Observation: An object of the topos $\mathrm{Set}^{\to}$ can be regarded as an indexed family (= formal coproduct) of sets $\{X_i\}_{i\in I}$. From this point of view, we have

$\{X_i\}_{i\in I}^{\{Y_j\}_{j\in J}} = \prod_{j\in J} \{X_i^{Y_j}\}_{i\in I}$.

We can generalize this observation. For a small category $C$, we can consider a new category $C^\triangleleft$ equipped with a new formal initial object. Finite ordinals are recursively obtained by $n+1 = n^{\triangleleft}$.
My claim is: $\mathrm{PSh}(C^\triangleleft)$ is the category of families of objects of $\mathrm{PSh}(C)$, and the exponentials in $\mathrm{PSh}(C^\triangleleft)$ is calculated by the same formula above with the exponentials in $\mathrm{PSh}(C)$.

(Here, we use not only the "exponential laws", but also the fact that the topos $\mathrm{PSh}(C^{\triangleleft})$ is stably locally connected. This ensures that the exponential functor ${-}^X\colon \mathcal{E}\to \mathcal{E}$ for a connected object $X$ preserves coproducts.)

This allows us to calculate exponentials in $\mathrm{PSh}(n)$ inductively.
Even for $\omega$, we can calculate exponentials with this formula from lower index $n\in \omega$. Exponential at index $n\in \omega$ does not depend on data of bigger indices $m>n$ since the forgetful functor $\mathrm{PSh}(\omega) \to \mathrm{PSh}(n)$ preserves exponentials (since this is a sheafification functor for an open subtopos).

This method get stuck for $\omega$ since $\omega^{\triangleleft}= \omega$.

If my intuition is correct, this should give us an inductive formula for $\mathrm{PSh}(\overrightarrow{n}^{\triangleleft}) \cong \mathrm{PSh}(\overrightarrow{n+1})$, yes?

For the base case, we have something like:

$${X_0}^{Y_0} = X^Y = \mathrm{Set}[X,Y]$$

and induction case, something like:

$${X_{n+1}}^{Y_{n+1}} = \sum_{h\in{\text{cod}(f_{n,n+1})}^{\text{cod}(g_{n,n+1})}} \prod_{i\in\text{cod}(g_{n,n+1})} f_{n,n+1}^{-1}(h(i))^{g_{n,n+1}^{-1}(i)}.$$

Where $f_{n,n+1}:: X_{n+1}\rightarrow X_n$ (resp. $g_{n,n+1}:: Y_{n+1}\rightarrow Y_n$).

(Here, I'm thinking of ${X_{n+1}}^{Y_{n+1}}$ as just a set.)

In other words, if I understand correctly, an exponential object $X^Y$ in $\mathrm{PSh}(\overrightarrow{n})$ is all possible ways to turn a $n-1$ length chain of composable functions named $X$ into the chain a $n-1$ length chain of composable functions denoted by the symbol $Y$, such that every that should commute in fact does, while itself being a $n-1$ length chain of composable functions.

Am I getting this right?

Ryuya Hora (Nov 13 2024 at 06:56):

Keith Elliott Peterson said:

If my intuition is correct, this should give us an inductive formula for $\mathrm{PSh}(\overrightarrow{n}^{\triangleleft}) \cong \mathrm{PSh}(\overrightarrow{n+1})$, yes?

Yes, I was just too lazy to write `\overrightarrow'.

Ryuya Hora (Nov 13 2024 at 07:34):

Keith Elliott Peterson said:

In other words, if I understand correctly, an exponential object $X^Y$ in $\mathrm{PSh}(\overrightarrow{n})$ is all possible ways to turn a $n-1$ length chain of composable functions named $X$ into the chain a $n-1$ length chain of composable functions denoted by the symbol $Y$, such that every that should commute in fact does, while itself being a $n-1$ length chain of composable functions.

Am I getting this right?

I could not understand what you mean. I'm sorry. (I could not understand what `named X' means.)

Let me clarify that a concrete description of the exponential object is already given by Kevin Carson:

Kevin Carlson said:

There's a general formula for exponentials in presheaf categories: $X^Y(c)=\mathrm{Set}^C(\hat c,X^Y)=\mathrm{Set}^C(Y\times \hat c,X)$, where $X,Y:C\to\mathrm{Set}, c$ is an object of $C,$ and $\hat c$ is the representable functor. For ordinals, the representables are either downward- or upward-closed chains and so $Y\times \hat c$ just erases one "end" or the other of $Y$;

This means that the set $(X^Y)_n \cong \mathrm{PSh}(\alpha)(Y\times n,X) \cong \mathrm{PSh}(\overrightarrow{n})(Y|_n,X|_n)$ (for any $\alpha\geq n$) is the set of ladders of height $n$:
スクリーンショット 2024-11-13 16.13.29.png
where $X|_n$ means the restriction of the functor to $\overrightarrow{n}$.

If $n=0$, this is just the usual exponential of sets. If $n=1$, this is the set of commutative diagram as in your original question.