### 1. Friends, Romans, ...

Extend your ears to me for I will now discuss the Kan extension condition. An important notion in topology is that of a “fibration”. Let us begin with two typical examples.

Consider the unit circle S1 as the space parametrising lines through the origin. Let M 2 × S1 be the “incidence locus” which consists of a point in 2 and a line which contains it; M is just another way of representing the Möbius strip. There is a natural map M S1. Any point on 2 other than the origin corresponds to a unique point in M; moreover, any line L in 2 that does not contain the origin maps bijectively to S1 -{L0} where L0 is the line parallel to L through the origin. Thus we see that M - L0 is isomorphic to the product of S1 -{L0} and a line.

Another example is the space Q of pairs of orthonormal vectors (v,w) (such that v.v = w.w = 0 and v.w = 1) in 3. We have a natural map Q S2 sending (v,w) to v. Given any fixed vector u, we can give a map from S2 -{u,-u} to Q by taking v to (v,(v × u)/|v × u|). More generally, as u varies on a great circle S1 (i. e. the intersection of S2 with a plane through the origin), we obtain a map S2 - S1 × S1 to Q.

1.1. Fibrations. In each case we have a map f : X Y such that each point y in Y has a neighbourhood U such that f-1(U) is homeomorphic to U × f-1(y). Such a map has traditionally been called a “fibre bundle”. However, the required of “locally product-like” is too restrictive since we are only studying spaces “upto homotopy”. From a topological point of view the important property of such a map is “path lifting” or more generally “homotopy extension”. Given any “test” space T and a closed subspace S, and a homotopy F : T × I Y , such that F0 : T Y “lifts” to G0 : T X, and the homotopy itself can be lifted along S, this lift can be extended to G : T × I Y . In other words, if we have a commutative diagram

where the left vertical arrow is the subspace inclusion, then there is a lift T ×I X. A map X Y of this kind has traditionally been called a “fibration” in algebraic topology.

Replacing topological spaces everywhere with simplicial complexes and the interval I with [1] we come up with the following notion of fibration of simplicial complexes. A map X Y of simplicial complexes is called a “fibration” if for any a complex K and any subcomplex L of K, and any diagram

we have a map K × [1] X that “completes” the diagram as above. Here e is either of the two vertices of [1]. A further technical requirement in this case is that X0 Y 0 must be surjective.

1.2. Reduction to basic cases. The above condition needs to checked for all complexes K and all sub-complexes L; this seems to be too much. Can we reduce this to a simpler condition? Suppose that is the smallest simplex of K that is not in L. Then the faces of lie in L so, using the map [n] K given by , we get a commutative diagram

Combining this with the above, we get a diagram

If can complete this diagram with a map [n] × [1] X, we can “add” to L and repeat the process. Now the last diagram is a particular case of the fibration diagram with K = [n] and L = [n]. Thus we see that an equivalent formulation of the fibration requirement is the completion of the last diagram and the surjectivity of X0 Y 0.

1.3. Even simpler. The subcomplex [n] ×{e} [n] × [1] of the simplicial complex [n] × [n] can be thought of as the boundary of a cylinder with one side capped contained in the solid cylinder. Another figure that is topologically similar is the inclusion of the boundary of a simplex with one face removed into the full simplex. Let k[N] denote the subcomplex of [n] in which the k-th face has been removed. The homotopy extension condition can be written as a lifting condition for the diagram:

Let X Y be a morphism of simplicial complexes. We say that it satisfies the Kan extension condition if for every such diagram we have a lifting [n] X which completes the diagram.

1.4. Equivalence. Let X Y be a morphism that satisfies the Kan extension condition. First of all note that 0[0] is the empty simplicial complex. Thus the Kan extension condition for n = 0 translates into the surjectivity of the map X0 Y 0. We wish to show that X Y is a fibration. By showing that a diagram of the form

We first note that [n] × [1] is the simplicial complex associated with the partially ordered set [0,n] × [0,1] where (i,j) < (k,l) if i < k and j < l. A non-degenerate k-simplex of this complex is associated with a strictly increasing sequence of k + 1 elements from this partially-ordered set. In particular, there are n + 1 n + 1-simplices, each is associated with a sequence like (0,0) << (i,0) < (i,1) << (n,1); we denote this simplex by i. Similarly, an n-simplex that lies in [n] × [1] is associated with a sequence of the type

or

Thus the two faces of i that are not in [n] × [1] are associated with the sequences (0,0) << (i - 1,0) < (i,1) << (n,1) and (0,0) << (i,0) < (i + 1,1) << (n,1). Depending on whether e is 0 or 1 the face dn+1n, or d00 respectively, is [n] ×{e,}. Let us assume that e is 0 as we can argue in a simlar fashion in the other case. The map is already defined on all faces of n except dnn. Thus we obtain a map n[n + 1] X which lifts the map [n + 1] Y that corresponds to the image of n in Y . By the Kan extension condition, we obtain a lifting [n + 1] X of n; in particular, we also obtain a lift of dnn. This latter is also dnn-1. Thus, we obtain a lift of the map given by n-1 to all its faces except dn-1n-1. This gives a map n-1[n + 1] X to which we again apply the Kan extension condition. Using the fact that dii-1 = dii, we can proceed by induction to lift each of the simplices i to X; hence we have the homotopy extension as required.

To prove the converse, we note that if X Y is a simplicial fibration then X0 Y 0 is a surjection by assumption. Thus we only need to prove the Kan extenion condition for simplices of dimension n + 1 where n > 0.

For each integer k between 0 and n there is an order preserving surjective map fk : [0,n] × [0,1] [0,n + 1]

Such an order preserving map induces a map of simplicial complexes [n] × [1] [n + 1] that we will denote by the same letter by abuse of notation. If is a strictly increasing sequence on [0,n] × [0,1] which contains at least one pair (t,0) and (t,1) for tk then the image of this sequence under fk is not strictly increasing; in other words, the image of the corresponding non-degenerate simplex is degenerate. In particular, the image of the n + 1-simplex l is non-degenerate only if l = k; the image of k is the non-degenerate n + 1-simplex in [n + 1]. Similarly, the image of an n-simplex that is contained in di[n] × [1] is non-degenerate only if it is dik for i < k and di+1k for i > k; the image of this n-simplex is the corresponding face of [n + 1]. Finally, the faces dkk and dk+1k are the images of [n] ×{1} and [n] ×{0} respectively.

To summarise, we have obtain a simplicial map fk : [n] × [1] [n + 1] which maps the [n] × [1] [n] ×{0} to k[n + 1] and maps [n] × [1] [n] ×{1} to k+1[n + 1]. Moreover, the only n + 1-simplex whose image is non-degenerate is k. Given a diagram

we obtain a diagram

If the map X Y has the homotopy extension property we obtain a map [n] × [1] X. Restricting this to k we obtain the required lifting [n + 1] X. We can argue similarly with k+1[n + 1]. We conclude that the Kan extension propery follows from the homotopy extension property as well.

Assignment.

1. Let F be a set and define a simplicial complex K by setting Kn = F and all maps to be identity. Show that this is a simplicial complex. We denote this simplicial complex by the same symbol F and call it the “constant” or discrete complex. Show that it is a Kan complex.
2. Let F be a set and define a simplicial complex K = E(F) by setting Kn = Fn+1 (F ×× F taken n times). Define the boundary map di : Fn+1 Fn as the map that “forgets” the i-th entry in the n+1-tuple. Define si : Fn Fn+1 as the map that duplicates the i-the entry. Show that this is a simplicial complex. Is it a Kan complex?