Lurie ([On the Classification of Topological Field Theories](https://arxiv.org/abs/0905.0465)), with some corrections by Calaque and Scheimbauer ([A note on the $(\infty,n)$-category of cobordisms](https://arxiv.org/abs/1509.08906)), famously constructed a symmetric monoidal $(\infty,n)$-category $\mathrm{Bord}\_n$ of $n$-dimensional smooth (co)bordisms.


Actually, the construction produces the following. Saying "such and such thing is an $(\infty,n)$-category" requires choosing a *model* of the $\infty$-category $\{\text{$(\infty,n)$-categories}\}$. By a "model" I mean a model category; "$(\infty,n)$-category" is then defined to mean fibrant-cofibrant object in the model. To foreshadow the question I'm going to get to, let me emphasize that a random object in the model category *is not* an $(\infty,n)$-category according to the model. Rather, it determines an $(\infty,n)$-category by (co)fibrant replacement.


Lurie and Calaque–Scheimbauer use the model of $(\infty,n)$-categories called "complete $n$-fold Segal spaces". The objects in the model are $n$-fold simplicial spaces, and the words "Segal" and "complete" refer to specific fibrancy conditions. Speaking approximately, "Segal" is a version of the requirement that composition be well-defined and single-valued: it rules out, for example, a situation where you have morphisms $f : X\to Y$ and $g : Y \to Z$, but no composition "$g\circ f$". "Completeness" has to do with invertibility: it says that a weak form of invertibility implies a strong form of invertibility. Fibrant replacement for the Segal condition goes through and adds in formal compositions of previously-incomposable pairs. Fibrant replacement for the completeness condition goes through and adds formal strong inverses to weakly-invertible things.


With this choice in mind, Lurie and Calaque–Scheimbauer build an $n$-fold simplicial space which deserves the name $\mathrm{Bord}\_n$ in the following sense: its entries are moduli spaces of bordisms. They prove that $\mathrm{Bord}\_n$ satisfies the Segal condition. However, they also emphasize that, for large enough $n$, **the $n$-fold simplicial space $\mathrm{Bord}\_n$ of Lurie and Calaque–Scheimbauer is not complete.** The issue has to do with the h-cobordism theorem, which can be used to give morphisms that I am calling "weakly but not strongly invertible".


For the purposes of stating the Cobordism Hypothesis, this in some sense "doesn't matter". First, as I foreshadowed above, $\mathrm{Bord}\_n$ has a completion $\widehat{\mathrm{Bord}}\_n$: when people talk about "the $(\infty,n)$-category of cobordisms" and cite Lurie or Calaque–Scheimbauer, they really mean $\widehat{\mathrm{Bord}}\_n$ and not $\mathrm{Bord}\_n$. Second, if $\mathcal{C}$ is any true $(\infty,n)$-category, i.e. a complete $n$-fold Segal space, then $\hom(\mathrm{Bord}\_n,\mathcal{C}) = \hom(\widehat{\mathrm{Bord}}\_n,\mathcal{C})$. So a universal-property characterization like "$\hom(\mathrm{Bord}\_n,\mathcal{C}) = \{\text{fully-dualizable objects in $\mathcal C$}\}$" will hold for both if it holds for either.


On the other hand, the replacement procedure to complete $\mathrm{Bord}\_n \leadsto \widehat{\mathrm{Bord}}\_n$ could be quite aggressive: in general, completion might require adding in lots of extra morphisms. For instance, one thing that I know that completion does is that it takes any h-cobordism $W$ between $M$ and $N$, slices it up $W$ parallel to the $M$- and $N$-boundaries, and then reads the slices as a new "smooth" family of manifolds connecting $M$ and $N$. In other words, this procedure goes through and potentially-aggressively modifies the moduli spaces of manifolds from what you originally thought they were.


The general version of my question is:



> 
> Just how aggressive is the completion $\mathrm{Bord}\_n \leadsto \widehat{\mathrm{Bord}}\_n$? How much does it add to the moduli spaces of $k$-dimensional bordisms?
> 
> 
> 


Here is a specific version of my question. The existence of h-cobordisms is related to the existence of *manifold factors*, which are non-manifold spaces $X$, like the [dogbone space](https://en.wikipedia.org/wiki/Dogbone_space), such that $X \times \mathbb R \cong \mathbb R^N$, or more generally $X \times M \cong N$ for some manifolds $M$ and $N$.



> 
> Do manifold factors show up as $k$-morphisms in $\widehat{\mathrm{Bord}}\_n$? Do fully-extended TFTs assign values to manifold factors?
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> 
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How aggressive is the fibrant replacement of $\mathrm{Bord}_n$?
The completeness condition is not really about making things invertible which weren't already. It is about where the information about invertible morphisms is stored.


We can already see this with $(\infty,1)$-categories of $n$-dimensional bordisms.


Since $\mathrm{Bord}\_n$ satisfies the Segal conditions, it makes sense to talk about compositions and invertibility of morphisms (points in the space of morphisms). What completeness would require is that if you have an invertible morphism (i.e. an invertible bordism), then it is already represented by a *path* in the *space of objects*.


In the bordism category (assuming we avoid dimension 4) the invertible morphisms are exactly the h-cobordisms.


In the Lurie and Calaque–Scheimbauer models, the space of objects is a union of $\operatorname{BDiff}(M)$, and the paths in this space correspond to bordisms which are mapping cylinders. In general there are h-cobordisms which are not mapping cylinders. So the "completeness fibrant replacement" would change the space to of objects to something like the classifying space of the subcategory of h-cobordisms.


But all of this is just about manifolds and you are not adding anything really new to the underlying category. You are just rearranging the furniture in the house. ;)


In particular manifold factors do not show up.


As a side note, the Lurie and Calaque–Scheimbauer models are also not Reedy fibrant (which is a condition even before Segal and complete). This means that for example there can be a morphism from $X$ to $Y$, and a path in the space of objects from $X$ to $X'$, but no morphism from $X'$ to $Y$.


