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 S^{1} as the space parametrising lines through the origin. Let
M ^{2} × S^{1} 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 S^{1}. 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 S^{1} -{L_{0}} where L_{0} is the line parallel to L through the origin.
Thus we see that M - L_{0} is isomorphic to the product of S^{1} -{L_{0}} 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 S^{2} sending (v,w) to v.
Given any fixed vector u, we can give a map from S^{2} -{u,-u} to Q by taking v to
(v,(v × u)/|v × u|). More generally, as u varies on a great circle S^{1} (i. e. the
intersection of S^{2} with a plane through the origin), we obtain a map S^{2} - S^{1} × S^{1} 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 F_{0} : T Y
“lifts” to G_{0} : 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 X_{0} 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
X_{0} 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 X_{0} 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
d_{n+1}_{n}, or d_{0}_{0} 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 d_{n}_{n}. 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
d_{n}_{n}. This latter is also d_{n}_{n-1}. Thus, we obtain a lift of the map given by _{n-1} to all
its faces except d_{n-1}_{n-1}. This gives a map _{n-1}[n + 1] X to which we again apply
the Kan extension condition. Using the fact that d_{i}_{i-1} = d_{i}_{i}, 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 X_{0} 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
f^{k} : [0,n] × [0,1] [0,n + 1]

To summarise, we have obtain a simplicial map f^{k} : [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.

- Let F be a set and define a simplicial complex K by setting K
_{n}= 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. - Let F be a set and define a simplicial complex K = E(F) by setting
K
_{n}= F^{n+1}(F ×× F taken n times). Define the boundary map d_{i}: F^{n+1}F^{n}as the map that “forgets” the i-th entry in the n+1-tuple. Define s_{i}: F^{n}F^{n+1}as the map that duplicates the i-the entry. Show that this is a simplicial complex. Is it a Kan complex?