Return to Answer

Probably not a full answer, but that could be starting point : at least it reduces the problem to a purely syntactical question... but I would assume that this syntactical question might have a nicer description...

Given a morphisms of Lawvere theory $S \to T$, I'll say it is co-étale (not sure this is a good terminology) if up to equivalence of the categories of models $T$ can be obtained from $S$ by only adding constant symbols and equation between constants.

In terms of models this implies that the forgetfull functor $T$-Mod $\to$ $S$-Mod identifies $T$-mod with a coslice of $S$-Mod. Obvisouly when this is the case the coslicing object is the initial model of $T$ seen as a model of $S$ by the forgetfull functor.

I claim that $T$ is co-extensive if and only if the two maps $\Delta: T \to T \times T$ and $T \to 1$ are co-étale,

where $T \times T$ denotes the Lawvere theory which at the level of underlying category is the products of $T$ with itself (and the set of sort is the product of the set of sort with itself) and whose models are pairs of models of $T$; and $1$ denotes the terminal (one sorted) lawevere theoy, which has only one model.

This follows from the fact that a category with coproduct and finite limits is extensive if and only if $C \times C \to C$ sending $(X,Y) \to X \coprod Y$ induces an equivalence $C \times C \simeq C /(1 \coprod 1)$ and if $C/0$ identifies with the terminal category.

Indeed, these conditions on $T$ would imply that the product functor $T$-Mod $\times$ $T$-Mod $\to T$-Mod identifies $T$-Mod $\times$ $T$-Mod with the coslice of $T$-Mod by the model $F_0 \times F_0$ (where $F_0$ is the initial model) which gives co-exentionality for binary product, and that the coslice of $T$-Mod by its terminal model is equivalent to the terminal category, which gives the extentionality condition of the terminal objects.

Note that the condition on $T \to T \times T$ is always a bit tricky as these obviously don't have have the same set of sorts, so that's never going to be the case that $T \times T$ can be obtained from $T$ by only adding constant and equation between constant. A big part of the story happen in the "up to equivalence of categories of models". Basically, you are going to enrich the theory of rings with two constant P_1 and P_2 that are orthogonal projections and you send one to zero and the other to $1$ in each component of the product. This does not make the underlying category of Ring$[P_1,P_2]$ equivalent to Ring $\times$ Ring, but they have the same Cauchy completion.

Probably not a full answer, but that could be starting point : at least it reduces the problem to a purely syntactical question... but I would assume that this syntactical question might have a nicer description...

Given a morphisms of Lawvere theory $S \to T$, I'll say it is co-étale (not sure this is a good terminology) if up to equivalence of the categories of models $T$ can be obtained from $S$ by only adding constant symbols and equation between constants.

In terms of models this implies that the forgetful functor $T$-Mod $\to$ $S$-Mod identifies $T$-mod with a coslice of $S$-Mod. Obviously when this is the case the coslicing object is the initial model of $T$ seen as a model of $S$ by the forgetful functor.

I claim that $T$ is co-extensive if and only if the two maps $\Delta: T \to T \times T$ and $T \to 1$ are co-étale,

where $T \times T$ denotes the Lawvere theory which at the level of underlying category is the products of $T$ with itself (and the set of sort is the product of the set of sort with itself) and whose models are pairs of models of $T$; and $1$ denotes the terminal (one sorted) Lawvere theory, which has only one model.

This follows from the fact that a category with coproduct and finite limits is extensive if and only if $C \times C \to C$ sending $(X,Y) \to X \coprod Y$ induces an equivalence $C \times C \simeq C /(1 \coprod 1)$ and if $C/0$ identifies with the terminal category.

Indeed, these conditions on $T$ would imply that the product functor $T$-Mod $\times$ $T$-Mod $\to T$-Mod identifies $T$-Mod $\times$ $T$-Mod with the coslice of $T$-Mod by the model $F_0 \times F_0$ (where $F_0$ is the initial model) which gives co-extensionality for binary product, and that the coslice of $T$-Mod by its terminal model is equivalent to the terminal category, which gives the extensionality condition of the terminal objects.

Note that the condition on $T \to T \times T$ is always a bit tricky as these obviously don't have have the same set of sorts, so that's never going to be the case that $T \times T$ can be obtained from $T$ by only adding constant and equation between constant. A big part of the story happen in the "up to equivalence of categories of models". Basically, you are going to enrich the theory of rings with two constant P_1 and P_2 that are orthogonal projections and you send one to zero and the other to $1$ in each component of the product. This does not make the underlying category of Ring$[P_1,P_2]$ equivalent to Ring $\times$ Ring, but they have the same Cauchy completion.

Probably not a full answer, but that could be starting point : at least it reduces the problem to a purely syntactical question... but I would assume that this syntactical answer might have a nice answer ?

Given a morphisms of Lawvere theory $S \to T$, I'll say it is co-étale (not sure this is a good terminology) if up to equivalence of the categories of models $T$ can be obtained from $S$ by only adding constant symbols and equation between constants.

In terms of models this implies that the forgetfull functor $T$-Mod $\to$ $S$-Mod identifies $T$-mod with a coslice of $S$-Mod. Obvisouly when this is the case the coslicing object is the initial model of $T$ seen as a model of $S$ by the forgetfull functor.

I claim that $T$ is co-extensive if and only if the two maps $\Delta: T \to T \times T$ and $T \to 1$ are co-étale,

where $T \times T$ denotes the Lawvere theory which at the level of underlying category is the products of $T$ with itself (and the set of sort is the product of the set of sort with itself) and whose models are pairs of models of $T$; and $1$ denotes the terminal (one sorted) lawevere theoy, which has only one model.

This follows from the fact that a category with coproduct and finite limits is extensive if and only if $C \times C \to C$ sending $(X,Y) \to X \coprod Y$ induces an equivalence $C \times C \simeq C /(1 \coprod 1)$ and if $C/0$ identifies with the terminal category.

Indeed, these conditions on $T$ would imply that the product functor $T$-Mod $\times$ $T$-Mod $\to T$-Mod identifies $T$-Mod $\times$ $T$-Mod with the coslice of $T$-Mod by the model $F_0 \times F_0$ (where $F_0$ is the initial model) which gives co-exentionality for binary product, and that the coslice of $T$-Mod by its terminal model is equivalent to the terminal category, which gives the extentionality condition of the terminal objects.

Note that the condition on $T \to T \times T$ is always a bit tricky as these obviously don't have have the same set of sorts, so that's never going to be the case that $T \times T$ can be obtained from $T$ by only adding constant and equation between constant. A big part of the story happen in the "up to equivalence of categories of models". Basically, you are going to enrich the theory of rings with two constant P_1 and P_2 that are orthogonal projections and you send one to zero and the other to $1$ in each component of the product. This does not make the underlying category of Ring$[P_1,P_2]$ equivalent to Ring $\times$ Ring, but they have the same Cauchy completion.

Probably not a full answer, but that could be starting point : at least it reduces the problem to a purely syntactical question... but I would assume that this syntactical question might have a nicer description...

Given a morphisms of Lawvere theory $S \to T$, I'll say it is co-étale (not sure this is a good terminology) if up to equivalence of the categories of models $T$ can be obtained from $S$ by only adding constant symbols and equation between constants.

In terms of models this implies that the forgetfull functor $T$-Mod $\to$ $S$-Mod identifies $T$-mod with a coslice of $S$-Mod. Obvisouly when this is the case the coslicing object is the initial model of $T$ seen as a model of $S$ by the forgetfull functor.

I claim that $T$ is co-extensive if and only if the two maps $\Delta: T \to T \times T$ and $T \to 1$ are co-étale,

where $T \times T$ denotes the Lawvere theory which at the level of underlying category is the products of $T$ with itself (and the set of sort is the product of the set of sort with itself) and whose models are pairs of models of $T$; and $1$ denotes the terminal (one sorted) lawevere theoy, which has only one model.

This follows from the fact that a category with coproduct and finite limits is extensive if and only if $C \times C \to C$ sending $(X,Y) \to X \coprod Y$ induces an equivalence $C \times C \simeq C /(1 \coprod 1)$ and if $C/0$ identifies with the terminal category.

Indeed, these conditions on $T$ would imply that the product functor $T$-Mod $\times$ $T$-Mod $\to T$-Mod identifies $T$-Mod $\times$ $T$-Mod with the coslice of $T$-Mod by the model $F_0 \times F_0$ (where $F_0$ is the initial model) which gives co-exentionality for binary product, and that the coslice of $T$-Mod by its terminal model is equivalent to the terminal category, which gives the extentionality condition of the terminal objects.

Note that the condition on $T \to T \times T$ is always a bit tricky as these obviously don't have have the same set of sorts, so that's never going to be the case that $T \times T$ can be obtained from $T$ by only adding constant and equation between constant. A big part of the story happen in the "up to equivalence of categories of models". Basically, you are going to enrich the theory of rings with two constant P_1 and P_2 that are orthogonal projections and you send one to zero and the other to $1$ in each component of the product. This does not make the underlying category of Ring$[P_1,P_2]$ equivalent to Ring $\times$ Ring, but they have the same Cauchy completion.

Probably not a full answer, but that could be starting point : at least it reduces the problem to a purely syntactical question... but I would assume that this syntactical answer might have a nice answer ?

Given a morphisms of Lawvere theory $S \to T$, I'll say it is co-étale (not sure this is a good terminology) if (up to equivalence) $T$ can be obtained from $S$ by only adding constant symbols and equation between constants.

In terms of models this implies that the forgetfull functor $T$-Mod $\to$ $S$-Mod identifies $T$-mod with a coslice of $S$-Mod. Obvisouly when this is the case the coslicing object is the initial model of $T$ seen as a model of $S$ by the forgetfull functor.

I claim that $T$ is co-extensive if and only if the two maps $\Delta: T \to T \times T$ and $T \to 1$ are co-étale,

where $T \times T$ denotes the Lawvere theory which at the level of underlying category is the products of $T$ with itself (and the set of sort is the product of the set of sort with itself) and whose models are pairs of models of $T$; and $1$ denotes the terminal (one sorted) lawevere theoy, which has only one model.

This follows from the fact that a category with coproduct and finite limits is extensive if and only if $C \times C \to C$ sending $(X,Y) \to X \coprod Y$ induces an equivalence $C \times C \simeq C /(1 \coprod 1)$ and if $C/0$ identifies with the terminal category.

Indeed, these conditions on $T$ would imply that the product functor $T$-Mod $\times$ $T$-Mod $\to T$-Mod identifies $T$-Mod $\times$ $T$-Mod with the coslice of $T$-Mod by the model $F_0 \times F_0$ (where $F_0$ is the initial model) which gives co-exentionality for binary product, and that the coslice of $T$-Mod by its terminal model is equivalent to the terminal category, which gives the extentionality condition of the terminal objects.

Probably not a full answer, but that could be starting point : at least it reduces the problem to a purely syntactical question... but I would assume that this syntactical answer might have a nice answer ?

Given a morphisms of Lawvere theory $S \to T$, I'll say it is co-étale (not sure this is a good terminology) if up to equivalence of the categories of models $T$ can be obtained from $S$ by only adding constant symbols and equation between constants.

In terms of models this implies that the forgetfull functor $T$-Mod $\to$ $S$-Mod identifies $T$-mod with a coslice of $S$-Mod. Obvisouly when this is the case the coslicing object is the initial model of $T$ seen as a model of $S$ by the forgetfull functor.

I claim that $T$ is co-extensive if and only if the two maps $\Delta: T \to T \times T$ and $T \to 1$ are co-étale,

where $T \times T$ denotes the Lawvere theory which at the level of underlying category is the products of $T$ with itself (and the set of sort is the product of the set of sort with itself) and whose models are pairs of models of $T$; and $1$ denotes the terminal (one sorted) lawevere theoy, which has only one model.

This follows from the fact that a category with coproduct and finite limits is extensive if and only if $C \times C \to C$ sending $(X,Y) \to X \coprod Y$ induces an equivalence $C \times C \simeq C /(1 \coprod 1)$ and if $C/0$ identifies with the terminal category.

Indeed, these conditions on $T$ would imply that the product functor $T$-Mod $\times$ $T$-Mod $\to T$-Mod identifies $T$-Mod $\times$ $T$-Mod with the coslice of $T$-Mod by the model $F_0 \times F_0$ (where $F_0$ is the initial model) which gives co-exentionality for binary product, and that the coslice of $T$-Mod by its terminal model is equivalent to the terminal category, which gives the extentionality condition of the terminal objects.

Note that the condition on $T \to T \times T$ is always a bit tricky as these obviously don't have have the same set of sorts, so that's never going to be the case that $T \times T$ can be obtained from $T$ by only adding constant and equation between constant. A big part of the story happen in the "up to equivalence of categories of models". Basically, you are going to enrich the theory of rings with two constant P_1 and P_2 that are orthogonal projections and you send one to zero and the other to $1$ in each component of the product. This does not make the underlying category of Ring$[P_1,P_2]$ equivalent to Ring $\times$ Ring, but they have the same Cauchy completion.