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1. • CommentRowNumber1.
   • CommentAuthorAndrew Stacey
   • CommentTimeOct 23rd 2012
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   Author: Andrew Stacey
   Format: MarkdownItexCan I check my working with someone who knows their cartesian closed categories better than I do (that's just about everyone, then)? Let $C$ be a cartesian closed category. I read on [[cartesian closed category]] that for an object $d \in C$ the functor $- \times d$ commutes with colimits. I'm pretty sure I understand this. So if I have a coequaliser diagram $a \rightrightarrows b \to c$ then $a \times d \rightrightarrows b \times d \to c \times d$ is also a coequaliser diagram. My context is (what else) [[generalised smooth spaces]]. I'm assuming that I'm working in a cartesian closed category. I'm also assuming that I have discrete (and indiscrete, but I don't think that matters here) smooth structures on a set (by inference from the name you can assume these to be left and right adjoint to the forgetful functor to $Set$ - for the moment I'm working with set-based theories). So I define the [[kinematic tangent space]] as the quotient of $C^\infty(\mathbb{R},X)$ by the relation $\alpha \simeq \beta$ if $\alpha(0) = \beta(0)$ and for all smooth $f \colon X \to \mathbb{R}$ then $(f \circ \alpha)'(0) = (f \circ \beta)'(0)$. Now's the first check-I-understand bit. I've been thinking of this as like a quotient in $Set$. But I guess I really ought to think of it as follows. Let $W$ be the **set** of pairs $(\alpha,\beta) \in C^\infty(\mathbb{R},X)$ satisfying the equivalence relation (so $\alpha(0) = \beta(0)$ and $(f\circ\alpha)'(0) = (f \circ\beta)'(0)$ for all $f$). Put the discrete smooth structure on this set (so it's a disjoint union of points) and define two maps $W \to C^\infty(\mathbb{R},X)$ to be the obvious maps. Then $T X$ is the coequaliser of $W \rightrightarrows C^\infty(\mathbb{R},X)$. First question: any smooth structure on $W$ such that the two maps were smooth would be fine here. Is that right? The use of the discrete one is simply that we know it will work. Thus $T X$ is a coequaliser. So as I'm in a cartesian closed category, the quotienting $C^\infty(\mathbb{R},X) \to T X$ commutes with products. In particular, if I took two spaces $X$ and $Y$ then $C^\infty(\mathbb{R}, X \times Y) \to T X \times T Y$ is a quotient map as it factors as: $$ C^\infty(\mathbb{R}, X \times Y) \cong C^\infty(\mathbb{R},X) \times C^\infty(\mathbb{R},Y) \to C^\infty(\mathbb{R},X) \times T Y \to T X \times T Y $$ and each one is a coequaliser map. At this stage I'm not sure if this is enough to conclude that $T(X \times Y) \cong T X \times T Y$. If not, though, I can show (from a little more careful examination) that the underlying set map of the natural map $T (X \times Y) \to T X \times T Y$ is a bijection, whereupon as both are quotients of the same space (and the natural map respects these quotient structures) the natural map must be an isomorphism. This is because my assumptions on the category (I think!) mean that quotients in it are formed by taking the set quotient and then imposing the finest smooth structure on it (this is a consequence of the existence of the indiscrete smooth structure, I believe). So since both have the same underlying set and the same quotient mapping they must have the same smooth structure. So ... how's my understanding? What have I missed?

   Can I check my working with someone who knows their cartesian closed categories better than I do (that’s just about everyone, then)?

   Let $C$ be a cartesian closed category. I read on cartesian closed category that for an object $d \in C$ the functor $- \times d$ commutes with colimits. I’m pretty sure I understand this. So if I have a coequaliser diagram $a \rightrightarrows b \to c$ then $a \times d \rightrightarrows b \times d \to c \times d$ is also a coequaliser diagram.

   My context is (what else) generalised smooth spaces. I’m assuming that I’m working in a cartesian closed category. I’m also assuming that I have discrete (and indiscrete, but I don’t think that matters here) smooth structures on a set (by inference from the name you can assume these to be left and right adjoint to the forgetful functor to $Set$ - for the moment I’m working with set-based theories). So I define the kinematic tangent space as the quotient of $C^\infty(\mathbb{R},X)$ by the relation $\alpha \simeq \beta$ if $\alpha(0) = \beta(0)$ and for all smooth $f \colon X \to \mathbb{R}$ then $(f \circ \alpha)'(0) = (f \circ \beta)'(0)$.

   Now’s the first check-I-understand bit. I’ve been thinking of this as like a quotient in $Set$. But I guess I really ought to think of it as follows. Let $W$ be the set of pairs $(\alpha,\beta) \in C^\infty(\mathbb{R},X)$ satisfying the equivalence relation (so $\alpha(0) = \beta(0)$ and $(f\circ\alpha)'(0) = (f \circ\beta)'(0)$ for all $f$). Put the discrete smooth structure on this set (so it’s a disjoint union of points) and define two maps $W \to C^\infty(\mathbb{R},X)$ to be the obvious maps. Then $T X$ is the coequaliser of $W \rightrightarrows C^\infty(\mathbb{R},X)$.

   First question: any smooth structure on $W$ such that the two maps were smooth would be fine here. Is that right? The use of the discrete one is simply that we know it will work.

   Thus $T X$ is a coequaliser. So as I’m in a cartesian closed category, the quotienting $C^\infty(\mathbb{R},X) \to T X$ commutes with products. In particular, if I took two spaces $X$ and $Y$ then $C^\infty(\mathbb{R}, X \times Y) \to T X \times T Y$ is a quotient map as it factors as:

   $$C^\infty(\mathbb{R}, X \times Y) \cong C^\infty(\mathbb{R},X) \times C^\infty(\mathbb{R},Y) \to C^\infty(\mathbb{R},X) \times T Y \to T X \times T Y$$

   and each one is a coequaliser map.

   At this stage I’m not sure if this is enough to conclude that $T(X \times Y) \cong T X \times T Y$.

   If not, though, I can show (from a little more careful examination) that the underlying set map of the natural map $T (X \times Y) \to T X \times T Y$ is a bijection, whereupon as both are quotients of the same space (and the natural map respects these quotient structures) the natural map must be an isomorphism. This is because my assumptions on the category (I think!) mean that quotients in it are formed by taking the set quotient and then imposing the finest smooth structure on it (this is a consequence of the existence of the indiscrete smooth structure, I believe). So since both have the same underlying set and the same quotient mapping they must have the same smooth structure.

   So … how’s my understanding? What have I missed?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeOct 23rd 2012
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   Author: Mike Shulman
   Format: MarkdownItex> any smooth structure on $W$ such that the two maps were smooth would be fine here. Well, any such smooth structure will suffice to let you *form* such a coequalizer, but at least *a priori*, different smooth structures might result in different coequalizers. However, if your category of smooth spaces has the property that any smooth function which is surjective on underlying sets is an epimorphism (such as when that the underlying-set functor is faithful), then you should get the same coequalizer whatever smooth structure you choose. Is that what you meant by "fine"? As for the rest of it, it might depend on what you mean by "quotient". Do you mean "coequalizer"? One potential issue is that it's not *a priori* true that the composite of coequalizers is a coequalizer. Perhaps you mean [[final lift]]? Is your category of smooth spaces a [[topological concrete category]]?

   any smooth structure on $W$ such that the two maps were smooth would be fine here.

   Well, any such smooth structure will suffice to let you form such a coequalizer, but at least a priori, different smooth structures might result in different coequalizers. However, if your category of smooth spaces has the property that any smooth function which is surjective on underlying sets is an epimorphism (such as when that the underlying-set functor is faithful), then you should get the same coequalizer whatever smooth structure you choose. Is that what you meant by “fine”?

   As for the rest of it, it might depend on what you mean by “quotient”. Do you mean “coequalizer”? One potential issue is that it’s not a priori true that the composite of coequalizers is a coequalizer. Perhaps you mean final lift? Is your category of smooth spaces a topological concrete category?

3. • CommentRowNumber3.
   • CommentAuthorAndrew Stacey
   • CommentTimeOct 23rd 2012
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   Author: Andrew Stacey
   Format: MarkdownItexThanks! So it looks like I'm missing some assumptions. > such as when that the underlying-set functor is faithful I'm happy assuming this. > Is your category of smooth spaces a topological concrete category? and this. Will that make it all work?

   Thanks! So it looks like I’m missing some assumptions.

   such as when that the underlying-set functor is faithful

   I’m happy assuming this.

   Is your category of smooth spaces a topological concrete category?

   and this.

   Will that make it all work?

4. • CommentRowNumber4.
   • CommentAuthorAndrew Stacey
   • CommentTimeOct 23rd 2012
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   Author: Andrew Stacey
   Format: MarkdownItex> As for the rest of it, it might depend on what you mean by “quotient”. I suspect that I didn't really have a clear picture in my head of what I did mean by "quotient". So let me see if I can come up with something that is rigorous. What I have in mind is that the kinematic tangent space is formed by "quotienting" the path space by some equivalence relation. So I want to take the *set* of equivalence classes of curves $\alpha$ where I set two to be the same if they satisfy the relation I said earlier and I want $C^\infty(\mathbb{R},X) \to T X$ to be a "quotient map". So let's *define* $T X$ as the coequaliser of $W \rightrightarrows C^\infty(\mathbb{R},X)$ as I described above. And my category is as nice as they come: topological concrete category. Certainly when I said "indiscrete" and "discrete" functors then I was thinking by analogy of indiscrete and discrete topologies.

   As for the rest of it, it might depend on what you mean by “quotient”.

   I suspect that I didn’t really have a clear picture in my head of what I did mean by “quotient”. So let me see if I can come up with something that is rigorous. What I have in mind is that the kinematic tangent space is formed by “quotienting” the path space by some equivalence relation. So I want to take the set of equivalence classes of curves $\alpha$ where I set two to be the same if they satisfy the relation I said earlier and I want $C^\infty(\mathbb{R},X) \to T X$ to be a “quotient map”.

   So let’s define $T X$ as the coequaliser of $W \rightrightarrows C^\infty(\mathbb{R},X)$ as I described above. And my category is as nice as they come: topological concrete category. Certainly when I said “indiscrete” and “discrete” functors then I was thinking by analogy of indiscrete and discrete topologies.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeOct 24th 2012
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   Author: Mike Shulman
   Format: MarkdownItexOkay, so in a topological concrete category, colimits are formed by making colimits in Set and equipping them with the final structure. So if "quotient map" means "final lift" then your coequalizer will be the set-coequalizer with the specified map being a quotient. And as you say, since products preserve coequalizers, they will preserve these quotient maps, and such quotient maps will be stable under composition. And then I think the rest of your argument should go through.

   Okay, so in a topological concrete category, colimits are formed by making colimits in Set and equipping them with the final structure. So if “quotient map” means “final lift” then your coequalizer will be the set-coequalizer with the specified map being a quotient. And as you say, since products preserve coequalizers, they will preserve these quotient maps, and such quotient maps will be stable under composition. And then I think the rest of your argument should go through.

6. • CommentRowNumber6.
   • CommentAuthorAndrew Stacey
   • CommentTimeOct 24th 2012
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   Author: Andrew Stacey
   Format: MarkdownItexYippee! Thanks for helping.

   Yippee! Thanks for helping.