Quite simply, I'd like to know what is the broadest or most natural context in which either (or both) of Mather's cube theorems hold. If you like, this may mean any of

  • What properties of $Top$ or $Top^*$ are essential to the proofs?
  • (where) are model/homotopical categories verifying Mather's theorems studied as such in the literature?
  • Are there more examples known verifying Mather's theorems?

I ask because Mather's proof strikes me as fairly gritty and seems to rely on explicit cellular constructions.

For reference, the cube theorems concern a cubical diagram whose faces commute up to homotopy in a coherent way, and assert

  1. If one pair of opposite faces are homotopy push-outs and the two remaining faces adjecent the source vertex are homotopy pull-backs, then the final two faces are also homotopy pull-backs
  2. If two pairs of opposite faces are homotopy pull-backs, and the remaining face adjacent the target vertex is a homotopy push-out, then the remaining face is a homotopy push-out.

Let $\mathcal{X}$ be an $\infty$-category (i.e., a homotopy theory) which admits small homotopy colimits, a set of small generators, and has the property that homotopy colimits in $\mathcal{X}$ commute with homotopy pullback. Then $\mathcal{X}$ satisfies the Mather cube theorem if and only if $\mathcal{X}$ is an $\infty$-topos: that is, it can be described as a left exact localization of an $\infty$-category of presheaves of spaces. (I learned this from Charles Rezk). Such homotopy theories are studied extensively in my book "Higher Topos Theory" (see in particular Proposition and the remark which follows it).

  • $\begingroup$ Jacob, I'm guessing that such things also have the following more general property: If $F$ and $G$ are functors $I\to \mathcal X$ such that for every $i$-morphism $i\to j$ you get a homotopy pullback square $(F(i)\to G(i))\to (F(j)\to G(j))$ then for every $i$-object $i$ you get a homotopy pullback square $(F(i)\to G(i))\to (hocolim F\to hocolim G)$. (The case of this when $I$ is the poset $a\leftarrow b\to c$ is the (first) Mather cube theorem.) $\endgroup$ – Tom Goodwillie Mar 3 '12 at 21:14
  • $\begingroup$ Or, Jacob, I suppose by "the Mather cube theorem" you mean statement 2 in the question. As I mentioned in my edit to my answer, statement 1 fails for the category of sets. $\endgroup$ – Tom Goodwillie Mar 3 '12 at 21:35
  • 1
    $\begingroup$ Hi Tom; I meant statement 1 (the category of sets is an ordinary topos, rather than an $\infty$-topos). Statement 2 would be a special case of the blanket hypothesis "homotopy colimits in X commute with homotopy pullback". The more general property you describe also holds in any $\infty$-topos. $\endgroup$ – Jacob Lurie Mar 4 '12 at 1:03
  • $\begingroup$ Oh my goodness wow! I'll have to keep studying these things. $\endgroup$ – some guy on the street Apr 16 '12 at 23:33

You might be interested in the work of Jean-Paul Doeraene: http://math.univ-lille1.fr/~doeraene/

In particular, on pages 8 and 9 of the paper Homotopy pull backs, push outs, and joins, he gives several examples of model categories satisfying the Cube Axiom, and identifies the cube maps in several which do not.

The motivation of much of this is to define and study Lusternik-Schnirelmann type invariants in model categories other than $Top$.


Here's a sketch proof of 2, sort of in the same spirit as Jeff Strom's answer:

These statements have equivalent formulations involving strictly commutative squares.

Denote a typical square by $\mathcal X$, with last space $X$ and two spaces $X_1$ and $X_2$ mapping into $X$, and $X_{12}$ mapping into both. It's called a homotopy pushout square if the resulting map from the homotopy pushout of $X_1\leftarrow X_{12}\to X_2$ to $X$ is a weak equivalence, and likewise for pullback.

Use the fact that if $X$ is the union of open sets $X_1$ and $X_2$ with intersection $X_{12}$ then the resulting square is a homotopy pushout.

Also use this converse: Any homotopy pushout square admits an equivalence from such an "open triad square"--a map that is a weak equivalence in all four corners.

Now let $\mathcal X\to \mathcal Y$ be a cube, a map of squares. Suppose that $\mathcal Y$ is a homotopy pushout square and that the four side squares are homotopy pullback squares.

Wlog the map $X\to Y$ is a fibration and the side squares are pullback (not just homotopy pullback) squares. Make an open triad square $\mathcal Y'$ and an equivalence $\mathcal Y'\to \mathcal Y$. Pull back to get a new square $\mathcal X'$. This is now also of the open triad kind, so it's a homotopy pushout square. And its map to $\mathcal X$ is an equivalence, so the latter is also a homotopy pushout square.

The other theorem, 1, implies 2 (for Top) but I don't think you can go the other way. And when I try to prove 1 for Top I end up using quasifibrations.

EDIT: The first theorem really seems deeper than the second. In the category of sets (with equivalence=iso, holim=lim, hocolim=colim) the second is true and the first is false.


The second cube theorem (if base and sides are ok, then so is the top) is a straightforward consequence of the formally crazy fact that if $p:E\to B$ is a fibration, $i:A\to B$ is a cofibration, and $p_A:E_A\to A$ is the pullback of $p$ along $i$, then $i:E_A\to E$ is also a cofibration.

  • 1
    $\begingroup$ That's in the category of simplicial sets, or what? $\endgroup$ – Tom Goodwillie Mar 2 '12 at 14:29
  • $\begingroup$ ahah! This will bear some thinking about. Tom, it's at least in $Top$-like things, which is (again) one of the likenesses I'm trying to get at; in this case, it seems to be the fact that ordinary pull-backs are subspaces of products in a very clean way. $\endgroup$ – some guy on the street Mar 2 '12 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.