The idea of a cofibration will be fundamental as we talk further about homotopy theory, as will its dual idea of a fibration. The point of the cofibration condition is the following. Oftentimes, we have a subspace and a map . We’d like to know when we can extend this map over all of . One useful criterion can be given by algebraic functors. For instance, singular homology can be used to show that the map does not extend over .

Many of the things we care about in algebraic topology are homotopy invariant, though. As a result, it would be nice to know when the question of how extends depends only on the homotopy class of . This is precisely the definition of a cofibration. Dualizing gives the definition of a fibration, which I will talk about some other time.

One can, not surprisingly, approach the business of fibrations and cofibrations in an axiomatic manner. This is the theory of model categories, due to Quillen. I don’t know anything about that though.

** Definition **

Definition 1A map is called acofibrationif whenever we have a space and a map as well as a homotopy such that

then we can “extend” the homotopy to such that

In general, one should think of as a subspace of to make life easier. Then the statement becomes simpler. Basically, given a map and a homotopy of , we can *extend* this homotopy to all of (or rather ). This actually doesn’t lose any generality provided we stick to Hausdorff spaces.

**So now on, assume all spaces are Hausdorff. We work with unpointed spaces; basepoints are not currently necessary.** In algebraic topology, one often restricts further to the category of “compactly generated” spaces, as developed in Steenrod’s paper *A convenient category of topological spaces* (LINk).

Let’s think about what the cofibration condition means in terms of a “universal example.” Let be a cofibration; I claim that is the inclusion of a closed subspace. Consider the union where is identified with . This is the so-called **mapping cylinder** of .

What I said about the mapping cylinder being universal is the following. To give a map and a homotopy compatible as in the definition of a cofibration is the same thing as homming out of the mapping cylinder. The universal such configuration is the map obvious inclusion and the map (which is not generally an inclusion, though it will be in the cofibration case). Clearly, there is a map

because and both map into . The homotopy extension property states precisely that there is a map

which is a half-inverse of . This implies that is a closed subspace, though.

Indeed:

Proposition 2Suppose is a map in the category of Hausdorff topological spaces that admits a retraction . Then is a closed subspace of .

*Proof:* Indeed, we can consider the (closed!) subspace consisting of such that . Clearly maps into because for any . The inverse map is just !

Without the Hausdorff condition, we would not know that was closed.

So “split injections” are closed immersions in the category of topological spaces.

Anyway, is a closed subspace of . In particular, is thus a subspace of under . This means that itself must be a closed immersion. We have proved the claim.

Conversely, it is easy to see that if

admits a retraction , then has the homotopy extension property. The reason is that if we have and a homotopy , we get a map out of the mapping cylinder and just use to get a map out of .

** The NDR condition **

It turns out, however, that the condition of being a cofibration is more than being a closed immersion. Basically, if is closed, then is a cofibration precisely when has a neighborhood which deformation retracts onto . This neighborhood, however, has to be expressible in a nice way.

Let us state this formally:

Definition 3A subspace of a Hausdorff space is called aneighborhood deformation retract (NDR)if there is a continuous with and a map such that, first, is just the projection and ; and, second, if , then

So basically the point is that determines the neighborhood that deformation retracts onto . If for all , then the condition is simply that deformation retracts onto and that be the zero set of a continuous real function.

Definition 4An NDR pair with function is called aDR pair (deformation retract)if everywhere.

The product of two NDR pairs is an NDR pair. We shall not prove this, as we need only a restricted version of this:

Proposition 5Suppose is a NDR pair and is a DR pair, then is a DR pair. In particular, is a deformation retract of .

*Proof:* We have the function which is zero precisely on . Moreover, we have the function which is zero precisely on and which is less than one everywhere. Moreover, we have the homotopies

These are the deformation retractions. Now we need to define such that is mapped as the identity and is mapped into .

So fix . The idea is that we push the first coordinate towards and push the second coordinate towards . But the naive approach of does not necessarily work. The problem is that this need not be stationary on . We have to modify this such that if one of the coordinates is very close to either or (but not the other), we don’t move the other much. Otherwise, something where the first coordinate in but the second is far from may get moved around a whole lot by the homotopy. And we want the homotopy to be stationary on .

In other words, we define

if (i.e. if is close to but is not necessarily close to ) but

if . If , we send this to

I claim that this is a deformation retraction of onto . Indeed, it is evident that has image in that subspace by the following argument. If , then the second coordinate of is , while if , then so that the first coordinate of , namely is in . Note that the function whose zero set is that subspace is just .

Finally, we can use this idea of an NDR pair to talk about cofibrations.

Proposition 6Let be a closed subspace of a Hausdorff space . Then the inclusion is a cofibration if and only if is a NDR pair.

*Proof:* Indeed, first suppose an NDR pair. Now is a DR pair by the obvious deformation retraction of onto the origin. Then we have shown that, by the previous proposition,

is a deformation retract. In particular, it is a retract. But this union is really just the mapping cylinder discussed earlier. So if the mapping cylinder is a retract of , then has the homotopy extension property and is a cofibration by the previous categorical discussion.

Conversely, suppose a cofibration. We prove that is an NDR pair. We know a priori that

admits a retraction.

First, we shall define the deformation retraction of a neighborhood of onto . The idea is to start at , then go up , then apply the retraction onto , then project down to the first coordinate. More precisely, we define via

where is the projection.

Now points near the top , intuitively, should be pulled by into and not far away to the bottom . So should be in if is near . We can make this formal by the following argument. Consider the open set ; this contains . If sends , then is an open set containing . If , then . So deformation retracts onto via .

We now need only check that there is a suitable . This is actually slightly tricky. What we do is

where is projection. If , clearly this is zero; conversely, if is such that has second coordinate for all , for . It follows that as well by continuity. So .

Since the is over , it is evident that if and only if , i.e. if does not when retracted fall all the way back down to the bottom slice . So is precisely the set defined above.

The continuity of is the hard part. For this, I would like to appeal to a general lemma. Once that is proved, we will have shown that is an NDR pair. So the lemma is the completion of the proof of this result.

The problem with is that it is a sup of a bunch of things. Taking the sup of infinitely many continuous functions generally doesn’t give you a continuous function. It does, however, under certain compactness conditions: if you have a “compact” family of functions.

Lemma 7Suppose is compact and is a continuous function. For each , consider the continuous function , . Then defines a continuous function on .

*Proof:* Indeed, for , the set is closed. This is because it is the intersection of closed sets

This is more generally the fact that the sup of a bunch of continuous functions is at least lower semicontinuous. The hard part will be to show that

is closed. For this, note that is closed as is continuous. Moreover, the sup is at least if and only if one of the ‘s is at least . What this means is that

if is the projection. However, the compactness of implies that is a *closed* map. That observation implies that is closed as well.

We have now proved the equivalence of cofibrations and NDR pairs.

I learned this mainly from Peter May’s *A Concise Course in Algebraic Topology*, available on his web page.

March 19, 2011 at 8:11 am

Great intro to cofibratons.

March 19, 2011 at 2:27 pm

Query. (No need to post) In definition 3 of this blog, should it be H(U x 1) rather than H(U x I)?

March 19, 2011 at 3:36 pm

There’s a typo in the definition of H. The sentence starting “In other words” should end with “if v(x) is less than u(x)”. (You reversed u and v and also reversed your inequality so the two reversals undid each other.)

More importantly, I am not convinced H is necessarily continuous. Assume there is a curve P(s) in X x Y (where P(s) is defined for s belongs to I) with the following properties:

(i) for all s other than s=1, u(P(s)) and v(P(s)) are non zero;

(ii) at s=1, u=v=0 (so in fact P(1) is in, and on the boundary of, A x B);

and

(iii)for s less than 1, u = 2v along P(s).

Let’s define p = P(1).

Clearly H(p,1) = p since u=v at p, but I don’t think this necessarily equals the limit of H(P(s),1). Therefore if such a curve existed, this would seem to lead to a potential discontinuity in H.

March 21, 2011 at 9:57 am

Hm. Er, ok, let me think about this when I get more time and get back to you.

April 3, 2011 at 9:51 pm

Thanks for the corrections! I finally got a chance to do this. I’m not sure I understand your objection, though. If we have such a curve, then as , the point is getting very close to both ; so applying shouldn’t move it very much, right?