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Is there a notion of "space" satisfying the following requirements?

  1. Spaces form (at least) a category; morphisms between spaces are called "continuous maps."

  2. Every topological space is a space, and the inclusion of $\mathbf{Top}$ in the category of spaces is fully faithful and injective on objects.

  3. The category $\mathbb{R}\mathbf{Mod}$ is a space.

  4. If $X$ is a topological space, a vector bundle over $X$ can be described as a continuous map $X \rightarrow \mathbb{R}\mathbf{Mod}.$

(I don't mind if some of these requirements are partially violated, they're just meant to be guidelines.)

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    $\begingroup$ Why the downvote? $\;\!$ $\endgroup$ – goblin Apr 30 '17 at 7:18
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    $\begingroup$ Topological stacks should fit the bill, more or less. Note that "injective on objects" is an evil condition. You can also consider the classifying space of the underlying groupoid of the topological category of f.d. vector spaces. $\endgroup$ – Qiaochu Yuan Apr 30 '17 at 7:58
  • $\begingroup$ Natural setting might be that of (Grothendieck) toposes. For any algebraic theory $T$ the category $BT$ of functors from finitely generated free $T$-models to sets is the classifying topos for flat $T$-models; that is, the category of geometric morphisms from $Shv(X)$ to $BT$ is equivalent to the category of sheaves of flat $T$-models on $X$. There is some work to do, though: one needs to get rid of infinitely generated flats to obtain in the remainder (locally) finitely generated projectives; second, $T$ must be the theory of modules not over a ring straightforwardly external to $X$. $\endgroup$ – მამუკა ჯიბლაძე Apr 30 '17 at 8:33
  • $\begingroup$ If one takes the sheaf of rings $C(-,\mathbb R)$ one ends up with $BT$ which is not independent of $X$. There must also be some way to take into account topology on $\mathbb R$ while still keeping it outside of $X$. Something like this has been achieved by Street in the very last paragraph of "Characterization of bicategories of stacks" $\endgroup$ – მამუკა ჯიბლაძე Apr 30 '17 at 8:34
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    $\begingroup$ @QiaochuYuan, I agree that topological stacks are the answer. So you should post it as an answer. (-: $\endgroup$ – Mike Shulman Apr 30 '17 at 8:37
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As Qiaochu said, probably what you want are topological stacks.

  1. Let $T$ be a small full subcategory of $\mathrm{Top}$, with the Grothendieck topology of open covers, and consider the 2-category of stacks of groupoids on $T$. Call its objects "spaces" and its morphisms "continuous maps". You could call its 2-cells "continuous transformations".

  2. Any topological space $X$ determines a sheaf $\mathrm{Top}(-,X)$ on $T$ and hence a stack. This functor is fully faithful on $T$, and often on a much larger subcategory of $\mathrm{Top}$. (For instance, if $T=\{\mathbb{R}^n\}$ then the functor is fully faithful on at least all topological manifolds.)

  3. The functor $T^{op} \to \mathrm{Gpd}$ defined by sending $X\in T$ to the groupoid of real vector bundles over $T$ is a stack, because vector bundles can be glued together over open covers. Call it your $\mathbb{R}\mathbf{Mod}$.

  4. If $X\in T$ (and often for many more $X\in\mathrm{Top}$), then by the Yoneda lemma, the groupoid of continuous maps and continuous transformations $\mathrm{Top}(-,X) \to \mathbb{R}\mathbf{Mod}$ is equivalent to $\mathbb{R}\mathbf{Mod}(X)$, i.e. the groupoid of real vector bundles on $X$.

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    $\begingroup$ It seems that the OP wants R-Mod to have also the non-invertible arrows. I don't know if this is possible in the framework of stacks, still seeing them as "spaces" (mathoverflow.net/questions/56962/… ? ). $\endgroup$ – Qfwfq May 2 '17 at 19:15
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    $\begingroup$ Okay; but you can of course consider stacks of categories too, and call them "spaces" if you like. (-: $\endgroup$ – Mike Shulman May 3 '17 at 0:05
  • $\begingroup$ Is there a way of doing this without presupposing the notion of a vector bundle? $\endgroup$ – goblin Oct 5 '17 at 5:32
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    $\begingroup$ @goblin Yes, $\mathbb{R}\mathbf{Mod}$ is the stackification of the constant prestack on the ordinary category of vector spaces, i.e. vector bundles are "exactly" what you get by "gluing together vector spaces along open covers". $\endgroup$ – Mike Shulman Oct 5 '17 at 5:39
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    $\begingroup$ @MikeShulman, that's awesome. $\endgroup$ – goblin Oct 5 '17 at 7:26
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NB As Qiaochu Yuan explains in the comment below, what follows is not correct: it only captures a very drastic quotient of the isomorphism groupoid of vector bundles; most likely - the groupoid of connected components of homs of the latter.

In a sense, there is a way to stay inside $\mathbf{Top}$. If I am not mistaken, the groupoid $[\![X,\coprod_nBO(n)]\!]$ with objects continuous maps $X\to\coprod_nBO(n)$ and morphisms homotopy classes of homotopies is equivalent to the groupoid of vector bundles on $X$ and their isomorphisms.

I don't know how to recover in this way vector bundle morphisms that are not isomorphisms, but OP does not mention this, so... :P

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  • $\begingroup$ This is not quite right. The groupoid of vector bundles and isomorphisms is a topological groupoid; I think what you want is to take its classifying space, which should produce a space homotopy equivalent to the mapping space from $X$ into that disjoint union. $\endgroup$ – Qiaochu Yuan May 2 '17 at 20:25
  • $\begingroup$ @QiaochuYuan What I describe is the path groupoid of $\coprod_nBO(n)^X$ (objects paths, morphisms homotopy classes of homotopies between paths). Do you mean this misses something? $\endgroup$ – მამუკა ჯიბლაძე May 2 '17 at 20:35
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    $\begingroup$ Yes. Consider the trivial vector bundle of rank $n$. Its automorphism group is the group $G$ of continuous maps from $X$ to $GL_n(\mathbb{R})$. On the other hand, the automorphism group of the trivial vector bundle, regarded as an object in the fundamental groupoid of $[X, BO(n)]$, is $\pi_0(G)$. $\endgroup$ – Qiaochu Yuan May 2 '17 at 22:53

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