I think that Proposition 2.36, together with Proposition 2.30 and the fact that coproducts are preserved, in these notes of Strickland establishes that pullbacks (and thus in particular equalisers) preserve colimits in the category of compactly generated weak Hausdorff spaces. I have not worked through the proof myself, though.

I too have not seen your approach taken elsewhere. Like many things, it is simple once one has seen it, but there are several little nuggets of creativity there that I feel one might easily overlook with regard to coming up with it in the first place, so I can believe that no-one has come up with it before!

]]>Richard, thank you! I’m not yet clear on “if equalizers preserve colimits”, though (i.e., how to turn that from hypothesis to fact). Perhaps I need to wake up more.

But if that works out easily, I think what I’d like to do is use that proof you gave in the beginning of #43, but at least archive the proof that’s there now and the other proof that was recently supplanted based on a similar suggestion of Karol.

(Perhaps the overall approach is original with me – I can affirm that I’ve not seen it elsewhere. But it would be odd if no one came up with it before, since it’s a simple idea piggybacking on a beautiful observation of Joyal.)

When I wake up more, I’ll be interested to read the rest of what you wrote.

]]>Continuing from the previous post…

Let $P$ be the product of $i_{!}i^{*}(X)$ and $i_{!}i^{*}(Y)$ in $\mathsf{Set}^{\Delta^{op}} / i_{!}i^{*}( \Delta^{0})$, that is to say, the pullback of the morphisms $i_{!}i^{*}(e_{X}) : i_{!}i^{*}(X) \rightarrow i_{!}i^{*}(\Delta^{0})$ and $i_{!}i^{*}(e_{X}) : i_{!}i^{*}(X) \rightarrow i_{!}i^{*}(\Delta^{0})$, where $e_{X}$ and $e_{Y}$ are the canonical morphisms coming from the fact that $\Delta^{0}$ is a final object of $\mathsf{Set}^{\Delta^{op}}$. Let $q_{X} : P \rightarrow i_{!}i^{*}(X)$ and $q_{Y} : P \rightarrow i_{!}i^{*}(Y)$ be the structural morphisms exhibiting $P$ as a pullback.

There is a canonical morphism $p : i_{!}i^{*}(X \times Y) \rightarrow P$ coming from the morphisms $i_{!}i^{*}(p_{X}) : i_{!}i^{*}(X \times Y) \rightarrow i_{!}i^{*}(X)$ and $i_{!}i^{*}(p_{Y}) : i_{!}i^{*}(X \times Y) \rightarrow i_{!}i^{*}(Y)$ , where $p_{X}$ and $p_{Y}$ are the structural morphisms exhibiting $X \times Y$ as product.

To demonstrate that $p$ is an isomorphism, as we require, it suffices, because $i^{*}$ is conservative, to show that $i^{*}(p)$ is an isomorphism. Because $i^{*}$ preserves limits (it is a right adjoint) , we have that $i^{*}(p_{X})$ and $i^{*}(p_{Y})$ exhibit $i^{*}(X \times Y)$ as the product of $i^{*}(X)$ and $i^{*}(Y)$. The morphisms $i^{*}(\epsilon_{X}) \circ i^{*}(q_{X})$ and $i^{*}(\epsilon_{Y}) \circ i^{*}(q_{Y})$ give rise to a canonical morphism $q : i^{*}(P) \rightarrow i^{*}(X \times Y)$, where $\epsilon : i_{!} i^{*} \rightarrow id$ is the co-unit for the adjunction between $i_{!}$ and $i^{*}$.

We now verify that $i^{*}(p)$ and $i^{*}i_{!}(q) \circ \eta_{i^{*}(P)}$ are inverse to one another, where $\eta : id \rightarrow i^{*}i_{!}$ is the unit of the adjunction between $i^{!}$ and $i^{*}$. To show that $i^{*}(p) \circ i^{*}i_{!}(q) \circ \eta_{i^{*}(P)}$ is the identity, we use the universal property of $i^{*}(P)$ as a pullback (appealing to the fact that $i^{*}$ preserves pullbacks, being a right adjoint). I’m sure that anybody reading could complete this part of the argument without my giving the details, but I’ll do so in case it saves anybody some time.

1) We have that:

$i^{*}(q_{X}) \circ i^{*}(p) \circ i^{*}i_{!}(q) \circ \eta_{i^{*}(P)}$

= $i^{*}\big( q_{X} \circ p \circ i_{!}(q) \big) \circ \eta_{i^{*}(P)}$

= $i^{*}\big(i_{!}i^{*}(p_{X}) \circ i_{!}(q) \big) \circ \eta_{i^{*}(P)}$

= $i^{*}\Big( i_{!}\big( i^{*}(p_{X}) \circ q \big) \Big) \circ \eta_{i^{*}(P)}$

= $i^{*}\Big( i_{!} \big( i^{*}(\epsilon_{X}) \circ i^{*}(q_{X}) \Big) \circ \eta_{i^{*}(P)}$

= $i^{*}i_{!}i^{*}(\epsilon_{X}) \circ i^{*}i_{!}i^{*}(q_{X}) \circ \eta_{i^{*}(P)}$

= $i^{*}i_{!}i^{*}(\epsilon_{X}) \circ \eta_{i^{*}i_{!}i^{*}(P)} \circ i^{*}(q_{X})$

= $id \circ i^{*}(q_{X})$

= $i^{*}(q_{X}).$

The penultimate equality comes from one of the triangles which commute as a consequence of the fact that $i_{!}$ and $i^{*}$ are adjoint.

2) By an entirely analogous argument, we have that $i^{*}(q_{Y}) \circ i^{*}(p) \circ i^{*}i_{!}(q) \circ \eta_{i^{*}(P)} = i^{*}(q_{Y})$.

3) We have that $i^{*}(q_{X}) \circ id = i^{*}(q_{X})$ and that $i^{*}(q_{Y}) \circ id = i^{*}(q_{Y})$.

4) We conclude from 1), 2), 3), and the universal property of $i^{*}(P)$ that $i^{*}(p) \circ i^{*}i_{!}(q) \circ \eta_{i^{*}(P)} = id$, as required.

To show that $i^{*}i_{!}(q) \circ \eta_{i^{*}(P)} \circ i^{*}(p)$ is the identity, we use the universal property of $i^{*}(X \times Y)$ as a product. The details are as follows.

1) We have that:

$i^{*}(p_{X}) \circ q \circ i^{*}(p)$

= $i^{*}(\epsilon_{X}) \circ i^{*}(q_{X}) \circ i^{*}(p)$

= $i^{*}(\epsilon_{X}) \circ i^{*}(q_{X} \circ p)$

= $i^{*}(\epsilon_{X}) \circ i^{*}i_{!}i^{*}(p_{X})$

= $i^{*}\big(\epsilon_{X} \circ i_{!}i^{*}(p_{X}) \big)$

= $i^{*}(p_{X} \circ \epsilon_{X \times Y})$

= $i^{*}(p_{X}) \circ i^{*}(\epsilon_{X \times Y}).$

2) By an entirely analogous argument, we have that $i^{*}(p_{Y}) \circ q \circ i^{*}(p) = i^{*}(p_{Y}) \circ i^{*}(\epsilon_{X \times Y})$.

3) We deduce from 1), 2), and the universal property of $i^{*}(X \times Y)$ as a product, that $q \circ i^{*}(p) = i^{*}(\epsilon_{X \times Y})$.

4) We conclude that

$i^{*}i_{!}(q) \circ \eta_{i^{*}(P)} \circ i^{*}(p)$

= $\eta_{i^{*}(X \times Y)} \circ q \circ i^{*}(p)$

= $\eta_{i^{*}(X \times Y)} \circ i^{*}(\epsilon_{X \times Y})$

= $id$,

as required. The second equality is a consequence of 3), and the final equality again comes from one of the triangles which commute as a consequence of the fact that $i_{!}$ and $i^{*}$ are adjoint.

We are done! We have demonstrated that $i^{*}(p)$ is an isomorphism, from which we conclude that $p$ is an isomorphism.

I would think that exactly the same kind of argument will work to show that $i_{!}i^{*}$ moreover preserves pullbacks, when viewed as a functor into $\mathsf{Set}^{\Delta^{op}} / i_{!}i^{*}(\Delta^{0})$, but have not gone through this.

]]>The argument I had in mind is as follows. Suppose that we have a pair of maps $f, g : X \rightarrow Y$ in $\mathsf{Set}^{\Delta^{op}}$, where $X$ is representable. Let $E$ be an equaliser of this pair of maps, and let $e : E \rightarrow X$ be the canonical map. Since $X$ is $n$-truncated and levelwise finite, so is $E$. Thus $E$ is clearly a finite colimit of representables. Hence $\left| E \right|$ is compact. Let $E'$ be the equaliser of $\left| f \right|$ and $\left| g \right|$ in (an appropriate variant of) $\mathsf{Top}$. Let $\mathsf{can} : \left| E \right| \rightarrow E'$ be the canonical map. We know, because $U \circ \left| - \right|$ preserves finite limits by the classifying topos argument, that $U(\mathsf{can})$ is an isomorphism of sets (where $U$ is the forgetful functor from $\mathsf{Top}$ to $\mathsf{Set}$). Moreover, $E'$ is Hausdorff, because it is a subspace of $\left| X \right|$, which is certainly Hausdorff. In summary, we have that $\mathsf{can}$ is a continuous bijection from a compact space to a Hausdorff space. Thus it is a homeomorphism, as required.

If equalisers preserve colimits in our variant of $\mathsf{Top}$, then we can reduce preservation of arbitrary equalisers $f, g : X \rightarrow Y$ to preservation of equalisers where $X$ is representable, to which the argument I have just given applies.

I would say that this argument is complementary to the one currently on the nLab: it is, I would say, a little simpler, but relies on stronger assumptions about $\mathsf{Top}$.

I agree with your concluding remarks and I think that the overall approach you have come up with is nice, and as about as optimal as it can be, even if some particular bits can be carried out in different ways. I guess it is correct to attribute the approach to you? The use of the characterisation of $\mathsf{Set}^{\Delta^{op}}$ is of course an old idea, due to Joyal I believe, but I mean the way to get from this to a result in (a variant of) $\mathsf{Top}$.

For David and anybody else interested: I realised that my argument is #36 is not quite correct. The thickened (I will use this in preference to ’fat’) geometric realisation is given by $\left| - \right| \circ i_{!} \circ i^{*}$, not $\left| - \right| \circ i_{*} \circ i^{*}$ as I wrote (too quickly!).

As penance, I will explain how to give a proof of (part of) Proposition 9 on the nLab page, namely that the thickened geometric realisation preserves products if we map to $\mathsf{Top} / \left|| \ast \right||$.

What I will actually do is show that if we view $i_{!} \circ i^{*}$ as a functor $\mathsf{Set}^{\Delta^{op}} \rightarrow \mathsf{Set}^{\Delta^{op}} / i_{!}i^{*}( \Delta^{0})$, then it preserves binary products. The part of Proposition 9 that concerns finite products is an immediate corollary. The argument I will give is purely abstract: it works for any adjunction for which $i^{*}$ is conservative, and this conservativity holds for any ’forgetful functor’ between presheaf categories of the kind we have here.

To be continued…

]]>Richard, if you see a way clear to simplifying the equalizer part of the argument here, I cordially invite you to do so. I’m very much interested in optimizing the argument, as so many texts seem to make the overall argument more difficult than I think it really is.

]]>A point perhaps worth mentioning is that, in a situation where one can reduce the case of equalisers to representables, one can use the same kind of argument as in #23, which I think is easier than the one showing that we have a subspace inclusion by appeal to (part of) cofibration generation. In this case, the fact that we have a finite colimit of representables is more or less immediate.

]]>Yes, I guess that, viewing $\Delta$ as a discrete $\mathsf{Top}$-enriched category, the category of simplicial topological spaces is the free $\mathsf{Top}$-enriched co-completion of $\Delta$, as long as we have a cartesian closed version of $\mathsf{Top}$. And, if so, then I think that indeed the same proof as for simplicial sets goes through for products.

For equalisers, if one can reduce to representables (as apparently is possible for compactly generated weak Hausdorff spaces), then the same argument as for simplicial sets seems to go through.

In short, I think I agree! And I think that, using $\mathsf{Top}$-enriched left and right Kan extensions, the argument that I gave to show that the other geometric realisation functor preserves pullbacks will go through, as I suggested in #38.

]]>Looking at the proof that products are preserved, if we assume we are in a category of spaces where colimits and pullbacks commute (i.e. coequalisers and pullbacks commute), then I think once we know that the map $||\Delta[m] \times \Delta[n]|| \to ||\Delta[m]|| \times_{||\Delta[0]||}||\Delta[n]||$ is an isomorphism we can just proceed as written, mutatis mutandis.

]]>Without having checked anything or thought about it in any detail, I would guess that, using enriched category theory, one could give essentially the same argument in the case of simplicial topological spaces. In other words, I would guess that one can reduce in the same way to showing that the ordinary geometric realisation functor from simplicial topological spaces to $\mathsf{Top}$ preserves pullbacks. It is particularly this that I have not thought about.

]]>Thanks, Richard. Yes, it’s the case of simplicial spaces that most of interest. But your argument for simplicial sets is very nice!

]]>Regarding #32: I think that it is straightforward to see that it is the case for simplicial sets rather than simplicial topological spaces. Indeed, consider the functor $\Delta_{\mathsf{semi}} \rightarrow \Delta$. This functor gives rise to a functor $i^{*} : \mathsf{Set}^{\Delta^{op}} \rightarrow \mathsf{Set}^{\Delta_{\mathsf{semi}}^{op}}$, a functor $i_{*} : \mathsf{Set}^{\Delta_{\mathsf{semi}}^{op}} \rightarrow \mathsf{Set}^{\Delta^{op}}$ which is right adjoint to it, and a functor $i_{!} : \mathsf{Set}^{\Delta_{\mathsf{semi}}^{op}} \rightarrow \mathsf{Set}^{\Delta^{op}}$ which is left adjoint to $i^{*}$. It seems to me that the geometric realisation functor you refer to (I can’t stand the name of it!) is $\left| - \right| \circ i_{*} \circ i^{*}$, where $\left| - \right|$ is the usual geometric realisation functor from $\mathsf{Set}^{\Delta^{op}}$ to $\mathsf{Top}$ (or whatever variant of this category you choose). Since both $i^{*}$ and $i_{*}$ are right adjoints, they both preserve pullbacks, so the question reduces to showing that the usual geometric realisation functor $\left| - \right|$ preserves pullbacks.

I have not thought about the case of simplicial topological spaces.

]]>Regarding #31…true, but, as in #21:

1) changing $\mathsf{Top}$ to the category of $k$-spaces, or whatever, doesn’t change the point I’m making regarding having a conceptual proof;

2) it does hold most of the time for $\mathsf{Top}$ itself, in particular in most cases that occur in practise.

]]>My guess is that it should be not *too* hard to show that the fibres of the map in #32 are compact, so we’d need to show that it was closed.

The only reference I’m aware of is a claim without proof in an unpublished preprint of Henriques and Gepner.

]]>David, I don’t know – I don’t have the requisite experience with fat realization. Maybe it’s time I sat down with it.

]]>@Todd - I’ve been asked twice for a proof recently that *fat* geometric realisation preserves pullbacks. Would your argument adapt to this case? I would guess that instead of $|\Delta[m] \times \Delta[n]|$ being compact, we would show that $||\Delta[m] \times \Delta[n]|| \to ||\Delta[0]||$ is proper, and so on.

(And, of course, it *doesn’t* actually hold for $Top$ itself; one has to modify it.)

Okay. I would tend to interprete the phrase “conceptual reason”, as used in your #24, to refer to “a reason to believe”.

]]>Karol: well, you’re right; thanks! And the same observation is already there in the proof, but you just condensed it for me. So maybe I’ll replace the proof with what you just told me.

Perhaps I was weighed down by the tradition of analyzing the simplicial structure of prisms, which seems to be part and parcel of just about every other proof I’ve seen that realization is left exact. Is prisms the word I want? I mean products of simplices.

]]>Todd, regarding your Lemma 2. If all that you need to conclude is that $|\Delta[m] \times \Delta[n]|$ is compact, then it suffices to observe that $\Delta[m] \times \Delta[n]$ has finitely many non-degenerate simplices. That’s clear since non-degenerate $k$-simplices in the nerve of a poset $P$ are exactly injective order preserving maps $[k] \to P$.

]]>I think that what you are giving are reasons to *believe* that the result holds for $\mathsf{Top}$, or that *suggest* that the result holds for $\mathsf{Top}$. What I am speaking of is a conceptual *proof* that it holds for $\mathsf{Top}$ itself, that it is to say a conceptual explanation for why it *actually is* true for $\mathsf{Top}$.

$Top$ is very close to being a quasitopos, and many quasitoposes embed in toposes (and one can even argue directly about geometric morphisms between quasitopoi). So it seems perfectly conceptual to me.

]]>As an aside, it is possible to show that the model structure on cubical sets with diagonals is Quillen equivalent to the one on unadorned cubical sets. Bas Spitters has a proof of this which I think is not available publically yet. I explained a proof to Bas over email a couple of weeks ago, using techniques which are similar to those I used above to show that the model structure on cubical sets with diagonals is Quillen equivalent to the one on simplicial sets.

If one combines the Quillen equivalence between simplicial sets and cubical sets with diagonals with one between the latter and unadorned cubical sets, we obtain a Quillen equivalence between simplicial sets and cubical sets. This is known by work of Cisinski, but the proof that I am suggesting is simpler (at least in my opinion!).

]]>Regarding #22: I certainly agree that this is a conceptual reason for geometric realisation out of simplicial sets into a topos to preserve products. But $\mathsf{Top}$ is not a topos, and I wouldn’t agree at all that pretending that it *is* a topos is a conceptual reason for why the geometric realisation out of simplicial sets into it preserves products.

I agree it holds for conceptual reasons that composing the geometric realisation functor with the forgetful functor to sets gives a functor which preserves finite limits. But we still have to go from this back to $\mathsf{Top}$, and it is this part that I do not regard as conceptual.

I do think that the approach that Todd describes in #23 is about as optimal as possible in this regard. Surely there is a simpler and more conceptual proof of Lemma 2, which is intuitively obvious, though? We could observe that:

i) any levelwise finite $n$-truncated simplicial set is a finite colimit of representables (one representable for every $m$-simplex, for $0 \leq m \leq n$);

ii) the product of an $m$-truncated and an $n$-truncated simplicial set is $m+n$-truncated;

iii) a pair of representable simplicial sets are certainly levelwise finite and $n$-truncated.

Just now I can’t think of a simple proof of ii) that improves much on the proof of Lemma 2, though.

Here is a more conceptual argument. The finite co-completion of $\Delta$ has products. It is tautologically the case that any object of this category is a finite colimit of representables, thus in particular it is the case for a product of representables. Passing from the finite co-completion of $\Delta$ to its full co-completion, namely to $\mathsf{Set}^{\Delta^{op}}$, preserves finite colimits and products.

I’m not sure if the use of the finite co-completion of $\Delta$ really allows this argument to qualify as simple, though! I think that the argument goes through if one works with presheaves of finite sets instead of the finite co-completion, and if so, maybe that argument gives the best balance between simplicity and conceptuality.

Anyhow, even with a conceptual proof of Lemma 2, I wouldn’t describe the overall proof that Todd describes as conceptual, because we rely on certain rather specific facts about topological spaces that we are not fruitfully going to be able to abstract to many interesting categories. In other words, I think it is about as conceptual as possible, but that the result that the product of the (ordinary) geometric realisations to $\mathsf{Top}$ of representable simplicial sets just does not admit an entirely conceptual proof.

Now, one might say: well, just use a topos instead of $\mathsf{Top}$, such as the topological topos of Johnstone. To this I would say: the proof of the pudding is in the eating! That is to say: maybe this can work, but there are things that need to be pursued. Firstly: does the topological topos admit a proper model structure that is Quillen equivalent to the one on $\mathsf{Set}^{\Delta^{op}}$?

I actually think that this may be fairly straightforward to show, if one takes the notion of weak equivalence of an arrow in the topological topos to be one which is a weak equivalence in $\mathsf{Set}^{\Delta^{op}}$ after applying the singular complex functor $S$. Because I think the unit of the adjunction $X \rightarrow S \left| X \right|$ is exactly the same as the usual one where we use $\mathsf{Top}$, and we know that this is a weak equivalence. It follows formally that the co-unit is also a weak equivalence. And because one seems to be able to work with CW-complexes and so on in the topological topos much as one does in $\mathsf{Top}$, I would not be at all surprised if one of the usual proofs of the existence of the Serre model structure went through. I would not be surprised either if the topological topos admits a Hurewicz-type model structure.

All this said, though, and even if the approach with the topological topos can work, there are still things that I prefer about the approach I suggested. These principally come down to simplicity: there is some non-trivial work involved in showing that simplicial sets are a classifying topos for intervals; the notion of a classifying topos itself is much more involved than having a free cartesian category; and it is nice to be able to know that geometric realisation into categories which are not topoi, such as $\mathsf{Cat}$ to give a simple example, preserve products for conceptual reasons, and in such a way that one has to check very little.

On a different note, I did in fact use the ’classifying topos’ approach to show that the realisation functor from cubical sets with diagonals to sets preserves equalisers. So I do not regard this particular step in my argument to be as conceptual as I would like. But, as I suggested, I strongly suspect that there is a direct proof that this realisation functor preserves equalisers/pullbacks. It may even be a consequence of the universal property of a free cartesian category: the nLab page free cartesian category suggests that this is at least the case when we have a discrete category.

]]>The parts of geometric realization that I had a hand in writing were written in order to try to bring out the conceptual point that Mike mentioned (that the topos $Set^{\Delta^{op}}$ classifies intervals). The idea is to see the left exactness $Set^{\Delta^{op}} \to Space$, where “$Space$” is some convenient category of topological spaces, by exploiting the left exactness of its underlying set functor $Set^{\Delta^{op}} \to Set$, which is the underlying conceptual part exploiting the fact that $Set$ is a topos.

Part of the left exactness of $Set^{\Delta^{op}} \to Space$ is not sensitive to whether we use a convenient category $Space$ or just $Top$. In particular, the arguments at geometric realization show that equalizers are preserved regardless of which is used. The only “messiness” there is to check that realization takes monomorphisms to (closed) subspace inclusions; since equalizers in $Top$ are equalizers in $Set$ equipped with the subspace topology, that’s all that is needed.

The “messiness” of proving that realization preserves products can be confined to a very small part: that in simplicial sets $Set^{\Delta^{op}}$, a product of representables $\Delta(-, [m]) \times \Delta(-, [n])$ is a quotient of a finite coproduct of representables. (If someone can improve on my proof of Lemma 2, please let me know!) Once you accept this, then you know that

${|\Delta(-, [m]) \times \Delta(-, [n])|}$ is compact,

${|\Delta(-, [m])|} \times {|\Delta(-, [n])|}$ is Hausdorff,

the canonical map ${|\Delta(-, [m]) \times \Delta(-, [n])|} \to {|\Delta(-, [m])|} \times {|\Delta(-, [n])|}$ is a bijection at the underlying set level (that’s the conceptual part),

and so ${|\Delta(-, [m]) \times \Delta(-, [n])|} \to {|\Delta(-, [m])|} \times {|\Delta(-, [n])|}$, being a continuous bijection from a compact space to a Hausdorff space, is a homeomorphism.

Then the remainder of the proof of product-preservation proceeds by the usual yoga of coends and Day convolution and whatnot – that part is completely conceptual, and is where the cartesian closedness of $Space$ really enters.

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