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# The nine model category structures on the category of sets

Tom Goodwillie mentioned on MathOverflow the striking fact that there are precisely nine model category structures on the category of sets. My friend Toby Barthel and I had some fun working out what the nine model structures are, which pairs are connected by a zigzag of Quillen equivalences and what the minimal length of such a zigzag is in each case (here this just boils down to showing there is no direct Quillen equivalence in a couple of cases for which there is a zigzag of length two between two structures). I don’t expect anyone would want to read this write up: some for lack of interest in the result and others because they would prefer to work it out themselves… At any rate, I found it a very instructive exercise which as a byproduct provides simple examples of

• a pair of weak factorization systems of the form $$(C \cap W, F)$$ and $$(C, F \cap W)$$ for which $$W$$ does not satisfy 2 out of 3, and
• pair of model categories connected by a zigzag of Quillen equivalences but with no direct Quillen equivalence between them.

Let’s get started. To find all the model structures we will succesively find all

1. lifting problems that have solutions,
2. pairs of functions having the lifting property,
3. weak factorization systems (WFS), and finally,
4. pairs of WFS that give rise to a model structure.

## All lifting problems that have solutions

Given a lifting problem $\begin{array}{ccc} A & \xrightarrow{g} & X \\ {\scriptstyle\pi}\downarrow & & \downarrow{\scriptstyle\rho} \\ B & \xrightarrow{f} & Y \\ \end{array}$ it is convenient to regard the vertical map $$\pi : A \to B$$ as giving a decomposition of $$A$$ into a disjoint union of possibly empty fibres given indexed by $$B$$, and similarly for $$\rho : X \to Y$$. We’ll also think of the bottom map $$f : B \to Y$$ as prescribing to which fibre of $$\rho$$ the top map $$g : A \to X$$ must send each fibre of $$A$$.

Let’s figure out when such a lifting problem has a solution. If a solution did exist, the commutativity of the top triangle means that the top map must collapse each non-empty fibre of $$\pi$$ to a single point in $$X$$; furthermore, in case this is satisfied, the solution is uniquely determined on points of $$B$$ with non-empty fibres above them. To be able to extend the solution to points of $$B$$ with empty fibres, these points must go to points of $$Y$$ possesing non-empty fibres.

In summary:

A solution to the lifting problem above exists if and only if

• the top map $$g$$ collapses each non-empty fibre of $$\pi$$ to a single point in $$X$$, and
• every point in the image of the bottom map has a non-empty fibre.

## All pairs of functions having the lifting property

Now, we want to describe all pairs of functions $$\pi : A \to B$$ and $$\rho : X \to Y$$ for which every commuting square is a lifting problem with a solution. Well, there are two conditions that are needed for any given lifting problem to have a solution, so let’s see when we can find a commuting squares with one of them failing.

If both $$\pi$$ and $$\rho$$ have a fibre with at least two points, we can make a lifting problem where some non-empty fibre of $$\pi$$ is not collapsed to a single point; otherwise, this collapsing condition is satisfied for every lifting problem involving $$\pi$$ and $$\rho$$.

To make a lifting problem failing the second condition, namely, one where some point in $$B$$ goes to a point of $$Y$$ with an empty fibre, we need the following to be true:

• both $$\pi$$ and $$\rho$$ have an empty fibre (in particular $$\pi$$ has some fibre, i.e., $$B$$ is non-empty), and
• either $$A$$ is empty (i.e., all fibres of $$\pi$$ are empty) or $$X$$ is not empty (i.e., $$X$$ can accomodate the non-empty fibres of $$\pi$$).

So, by carefully de Morganizing the above, we’ve proved that $$\pi$$ has the left lifting property (henceforth, LLP) with respect to $$\rho$$ if and only if

• at least one of $$\pi$$ and $$\rho$$ is injective, and
• either at least one of $$\pi$$ and $$\rho$$ is surjective, or $$A \neq \emptyset = X$$.

Of course, any function $$\emptyset \to Y$$ is injective, so we can simplify the criterion to:

$$\pi$$ has the LLP with respect to $$\rho$$ if and only if

• $$A \neq \emptyset = X$$ (in this case there simply are no lifting problems involving $$\pi$$ and $$\rho$$), or
• out of $$\pi$$ and $$\rho$$, at least one is injective and at least one is surjective.

## All weak factorization systems

A little more notation: given a function $$\pi$$, let $$\pi^\perp$$ be the collection of all functions having the RLP with respect to $$\pi$$; and let $$\bar{\pi} := {}^\perp (\pi^\perp)$$, the collection of all functions having the LLP with respect to $$\pi^\perp$$.

A WFS is “(left) mono-generated” if it is of the form $$(\bar{\pi}, \pi^\perp)$$ for some function $$\pi : A \to B$$. A priori, not every $$\pi$$ need generate a WFS, only those do for which every function can be written as a composite of a function in $$\bar{\pi}$$ and one in $$\pi^\perp$$; but it will turn out that in the category of sets all functions do generate a WFS. It is straightforward to use the characterization of the previous section to find all mono-generated WFS on the category of sets, and once we’re in posession of the full list, it will be easy to show there is just one more WFS.

OK, now let’s go over all the functions $$\pi$$ and compute their $$\pi^\perp$$’s and $$\bar{\pi}$$’s. We’ll use the abreviations $${\text{bij}}, {\text{inj}}, {\text{surj}}$$ and $${\text{any}}$$ for the obvious classes of functions, $${{\text{inj}}_\emptyset}$$ for inclusions of the empty set into some other arbitrary set, and the subscript “$${\neq \emptyset}$$” to mean “with non-empty domain”. There are several cases to consider:

1. $$\pi$$ is a bijection. Then $$\pi^\perp = {\text{any}}$$ and $$\bar{\pi} = {\text{bij}}$$.
2. $$\pi$$ is an injection which is not surjective.
1. $$A = \emptyset$$. Then $$\pi^\perp = {\text{surj}}$$ and $$\bar{\pi} = {\text{inj}}$$.
2. $$A$$ is non-empty. Then $$\pi^\perp = {\text{surj}}\cup {{\text{inj}}_\emptyset}$$, and $$\bar{\pi} = {\text{inj}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}}$$.
3. $$\pi$$ is a surjection which is not injective. Then $$\pi^\perp = {\text{inj}}$$ and $$\bar{\pi} = {\text{surj}}$$.
4. $$\pi$$ is neither injective nor surjective. Then $$\pi^\perp = {\text{bij}}\cup {{\text{inj}}_\emptyset}$$ and $$\bar{\pi} = {\text{any}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}}$$.

It is easy to check that all of these pairs $$(\bar{\pi}, \pi^\perp)$$ do in fact produce WFS, and it is also clear by a disturbing lack of symmetry that at least more non-mono-generated WFS exists: $$({\text{any}}, {\text{bij}})$$. The above five plus this sixth one form a full list as we’ll argue now. Order the WFS by reverse inclusion of the right class (or equivalently, usual inclusion of the left class) to get the poset with the following Hasse diagram (where we’ve just indicated the right classes of the WFS for reasons which will soon be clear):

$\begin{array}{ccccc} & & {\text{bij}}& & \\ & \nearrow & & \nwarrow & \\ {\text{bij}}\cup{{\text{inj}}_\emptyset}& & & & {\text{surj}}\\ \uparrow & & \nwarrow & & \uparrow \\ {\text{inj}}& & & & {\text{surj}}\cup{{\text{inj}}_\emptyset}\\ & \nwarrow & & \nearrow & \\ & & {\text{any}}& & \\ \end{array}$

Now, given an arbitrary WFS $$(L,R)$$, we have $$R = \bigcap_{\pi \in L} \pi^\perp$$, so that $$R$$ is an intersection of the various $$\pi^\perp$$ we found before. However, one can easily check by looking at the Hasse diagram that with the addition of the new minimal right class $${\text{bij}}$$, the collection of the $$\pi^\perp$$ we found becomes closed under intersections. This shows that we have a complete list of WFS for the category of sets.

## All model structures

We will use the following characterization of model structure:

Three classes of morphisms $$W$$, $$C$$ and $$F$$ form a model structure if and only if $$W$$ satisfies 2 out of 3 and both $$(C \cap W, F)$$ and $$(C, F \cap W)$$ are WFS.

Now, if we have a model structure $$(W,C,F)$$, the two WFS associated to it automatically satisfy $$(C \cap W, F) \le (C, F \cap W)$$, so we can read off candidate pairs of WFS from the Hasse diagram in the previous section.

If the two WFS were equal, $$(C \cap W, F) = (C, F \cap W)$$, we’d have $$W = {\text{any}}$$ which does indeed satisfy 2 out of 3, guaranteeing we get a model structure. Thus, there are six model structures where $$W = {\text{any}}$$ and $$(C,F)$$ is any of the six WFS found above. Note that all of these have the terminal category as homotopy category.

Let’s deal now with model structures where $$(C \cap W, F)<(C, F \cap W)$$. Well, those two WFS must be among the comparable pairs in the poset described in the previous section. Given a candidate pair of WFS, how do we test if they come from a model category? We can:

1. Compute $$W = (F \cap W) \circ (C \cap W)$$. (By this we mean $$W$$ consists of all compositions of a function in $$F \cap W$$ and a function in $$C \cap W$$.)
2. Check consistency, i.e., that the supposed $$C \cap W$$ from the WFS actually is the intersection of $$C$$ and the newly computed $$W$$. (And similarly for $$F \cap W$$.)
3. Check that $$W$$ satisfies 2 out of 3.

OK, let’s get started:

1. $$(C \cap W, F) = ({\text{bij}}, {\text{any}})$$

In this case, we get $$W = (F \cap W) \circ (C \cap W) = F \cap W$$, since every class is contains all bijections and is closed under composition. Now, of all the right classes we have, only $${\text{any}}$$, and $${\text{bij}}$$ satisfy 2 out of 3. We’re assuming the WFS are different, so we need only get one new model structure:

• $$(C, F, W) = ({\text{any}}, {\text{any}}, {\text{bij}})$$.
2. $$(C, W \cap F) = ({\text{any}}, {\text{bij}})$$

Here, analogously to the previous case, we have $$W = C \cap W$$. Of all the left classes we have, only $${\text{bij}}$$, $${\text{any}}$$ and $$({\text{any}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}})$$ satisfy 2 out of 3. So we get one new model structure:

• $$(C, F, W) = ({\text{any}}, {\text{bij}}\cup {{\text{inj}}_\emptyset}, {\text{any}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}})$$.
3. There remains the case where both WFS are not maximal and not minimal. Looking at the Hasse diagram above, we see there are three pairs of such WFS:
1. $$(C \cap W, F) = ({\text{surj}}, {\text{inj}})$$ and $$(C, F \cap W) = ({\text{any}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}}, {\text{bij}}\cup {{\text{inj}}_\emptyset})$$. We get $$W = {\text{surj}}\cup {{\text{inj}}_\emptyset}$$ which does not satisfy 2 out of 3. This gives an example of a pair of WFS which are of the form $$(C \cap W,F)$$ and $$(C, F \cap W)$$ for three classes $$C$$, $$F$$, $$W$$, but which do not form a model structure.
2. $$(C \cap W, F) = ({\text{inj}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}}, {\text{surj}}\cup {{\text{inj}}_\emptyset})$$ and $$(C, F \cap W) = ({\text{any}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}}, {\text{bij}}\cup {{\text{inj}}_\emptyset})$$. We get $$W = {\text{inj}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}}$$ which does not satisfy 2 out of 3.
3. $$(C \cap W, F) = ({\text{inj}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}}, {\text{surj}}\cup {{\text{inj}}_\emptyset})$$ and $$(C, F \cap W) = ({\text{inj}}, {\text{surj}})$$. We get $$W = {\text{any}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}}$$, which does satisfy 2 out of 3 and in fact we get the last model structure:
4. $$(C, F, W) = ({\text{inj}}, {\text{surj}}\cup {{\text{inj}}_\emptyset}, {\text{any}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}})$$.

Let’s take inventory of the homotopy categories for the nine model structures: the first six we found, for which all functions are weak equivalences, have the terminal category for homotopy category; the two with weak equivalences given by $${\text{any}}{_{\neq\emptyset}}\cup {\{\text{id}_\emptyset\}}$$ have the discrete category on two objects (one object for the empty set and one for all other sets) as their homotopy category; and the one with bijections as weak equivalences has the category of sets as homotopy category. Notice that these are the homotopy category of (-2)-types (i.e., contractible spaces), (-1)-types (spaces that are either empty or contractible) and 0-types (disjoint unions of contractible components) respectively. Let no-one say that model structures on the category of sets are not about homotopy theory.

For convenient reference, here’s a table of the model structures:

Cofibrations     Fibrations Weak equivalences Homotopy category
$${\text{bij}}$$ $${\text{any}}$$ $${\text{any}}$$ (-2)-types
$${\text{surj}}$$ $${\text{inj}}$$ $${\text{any}}$$ (-2)-types
$${\text{inj}}{_{\neq\emptyset}}\cup{\{\text{id}_\emptyset\}}$$ $${\text{surj}}\cup{{\text{inj}}_\emptyset}$$ $${\text{any}}$$ (-2)-types
$${\text{any}}{_{\neq\emptyset}}\cup{\{\text{id}_\emptyset\}}$$ $${\text{bij}}\cup{{\text{inj}}_\emptyset}$$ $${\text{any}}$$ (-2)-types
$${\text{inj}}$$ $${\text{surj}}$$ $${\text{any}}$$ (-2)-types
$${\text{any}}$$ $${\text{bij}}$$ $${\text{any}}$$ (-2)-types
$${\text{inj}}$$ $${\text{surj}}\cup{{\text{inj}}_\emptyset}$$ $${\text{any}}{_{\neq\emptyset}}\cup{\{\text{id}_\emptyset\}}$$ (-1)-types
$${\text{any}}$$ $${\text{bij}}\cup{{\text{inj}}_\emptyset}$$ $${\text{any}}{_{\neq\emptyset}}\cup{\{\text{id}_\emptyset\}}$$ (-1)-types
$${\text{any}}$$ $${\text{any}}$$ $${\text{bij}}$$ 0-types

## Quillen equivalences

Finding all Quillen equivalences between these model structures is easy. First of all, we need only check whether the identity functor is a Quillen equivalence, since all self-equivalences of the category of sets are naturally isomorphic to the identity (a self-equivalence must preserve singletons, since they are terminal, and thus any other set $$S$$ since it is the coproduct of $$|S|$$ singletons). Thus two of these model structures will be Quillen equivalent if they have the same weak equivalences (we saw that different weak equivalences led to non-equivalent homotopy categories) and their classes of cofibrations are comparable (in the inclusion order).

So the two model structures for (-1)-types are Quillen equivalent. As for the six model structures for (-2)-types, that are of the form $$(C,W,{\text{any}})$$ where $$(C,W)$$ is one of the WFS, we see that two coming from comparable WFS will be Quillen equivalent. There are three incomparable pairs of WFS, and no pair of them is directly Quillen equivalent, since the identity functor is not Quillen in either direction. Thse pairs, whoever, are connected by a zigzag of Quillen equivalences of length 2, and even by two such: one passing through the maximal WFS in the partial order and one passing through the minimal WFS. Confusingly, in these zigzags, both functors are the identity —but one is a left Quillen functor and the other a right Quillen functor.