Last time, we explained the idea of a cofibration in terms of a useful homotopy extension property. We showed that, for Hausdorff spaces, cofibrations turn out always to be closed immersions. Moreover, we showed that a pair ${(X, A)}$ with ${A \subset X}$ closed is a cofibration precisely when it satisfies a technical condition of ${A}$ being a neighborhood deformation retract. However, we have yet to give useful examples. The main result in this post is that a relative CW complex (for instance, a CW pair) leads to a cofibration.

1. The mapping cylinder

It turns out that, up to homotopy equivalence, every map is a cofibration. The method of showing this is to use the mapping cylinder. So let ${f: X \rightarrow Y}$ be a map. Recall that the mapping cylinder ${M_f}$ is the quotient space ${(X \times I) \cup Y}$ where ${(x,1)}$ is identified with ${f(x) \in Y}$. We have an inclusion map ${X \rightarrow M_f}$ sending ${x \rightarrow (x,0)}$ and a projection map ${M_f \rightarrow Y}$ sending ${(x,t) \rightarrow f(x)}$.

The projection map ${M_f \rightarrow Y}$ is a homotopy equivalence. In fact, ${M_f}$ deformation retracts onto ${Y}$ because ${X \times I}$ deformation retracts onto ${X \times \left\{1\right\}}$. The homotopy inverse of the inclusion ${Y \rightarrow M_f}$ can be taken to be this projection. So we can factor ${f: X \rightarrow Y}$ via

$\displaystyle X \rightarrow M_f \rightarrow Y,$

where the second map is a homotopy equivalence.

Proposition 1 ${X \rightarrow M_f}$ is a cofibration.

Proof: Indeed, we can consider the function ${u: X \times I \cup Y \rightarrow I}$ sending ${(x,t) \rightarrow \min(2t, 1)}$ (and ${u(y)=1, y \in Y}$). Then ${u^{-1}(0) = X \times \left\{0\right\}}$. Moreover, it is easy to see that ${u^{-1}([0,1))}$ deformation retracts onto ${X = X \times \left\{0\right\}}$. We can actually extend the deformation retraction to all of ${X \times I \cup Y}$ by making it go slower on ${(x,t)}$ as ${t}$ gets larger, and eventually stop moving. For instance, we could choose

$\displaystyle H((x,t),t') = (x, t - t' (1-u(x,t))).$

It is thus easy to see that ${(X \times I \cup Y, X)}$ is an NDR pair, so the inclusion of ${X}$ in the mapping cylinder is a cofibration.

$\Box$

In particular, every map is equivalent in the homotopy category to a cofibration.

General facts

One of the key results one uses over and over again in homotopy theory is that a CW complex has the homotopy extension property with respect to any subcomplex. The reason is that it allows you to construct homotopies piece by piece.

More generally, we shall see this with a bit of categorical nonsense:

Proposition 2 Consider a cocartesian diagram of topological spaces

Then if ${A \rightarrow C}$ is a cofibration, so is ${B \rightarrow D}$.

Let us recall what the idea of a “cocartesian” diagram is. This means that ${D}$ is the push-out of ${B,C}$ with respect to ${A}$. In other words, ${D}$ is the disjoint sum ${B +C}$ with the images of ${A}$ in both identified. To hom out of ${D}$ is the same thing as homming out ${B, C}$ such that the pull-backs to ${A}$ are compatible.

Proof: The reason is purely formal. Suppose ${A \rightarrow C}$ has the homotopy lifting property and we have maps ${B \times I \rightarrow X}$, ${D \rightarrow X}$ which agree on ${B \times \left\{0\right\}}$. The first gives a map ${A \times I \rightarrow X}$; the second gives a map ${C \rightarrow X}$. These two agree on ${A = A \times \left\{0\right\}}$.

The homotopy lifting property says that we can extend ${A \times I \rightarrow X}$ and ${C \rightarrow X}$ to ${C \times I \rightarrow X}$. We thus have two maps ${C \times I \rightarrow X, B \times I \rightarrow X}$ which agree on ${A \times I}$. Now ${D \times I}$ is the push-out of ${C \times I, B \times I}$ with respect to ${A \times I}$ because push-outs (and more generally, quotients) commute with products with a locally compact Hausdorff space. (This is just one of those random facts in general topology that keep cropping up again and again.)

So since we have hommed out of ${C \times I , B \times I}$ into ${X}$, we get a map ${D \times I \rightarrow X}$. This extends ${C \times I \rightarrow X}$. We need to just check that extends ${D \rightarrow X}$. But this is evident because the restriction of ${D \times I \rightarrow X }$ to ${D \times \left\{0\right\}}$ is determined by the restrictions to ${B \rightarrow X, C \rightarrow X}$, and these were done appropriately. $\Box$

As a result, we can show that attaching a space by a cofibration leads to a cofibration. Since CW complexes are obtained by attaching cells, this is very nice.

Suppose ${Y}$ is a topological space. Let ${A \subset B}$ be a subspace; suppose we have a map ${f: A \rightarrow Y}$. Then we can consider the attached space

$\displaystyle Y \cup_f B =( Y + B)/\left\{a \sim f(a)\right\}.$

In particular, we tack ${B}$ onto ${X}$ and identify the points in ${A}$ with their images. The standard example of doing this is when ${A = S^{n-1} \subset D^n}$. Then ${Y \cup_f D^n}$ is said to be obtained from ${Y}$ by attaching ${n}$-cells. This process is how one can construct a so-called relative CW complex.

Proposition 3 Suppose ${(B, A)}$ is an NDR pair (i.e. ${A \hookrightarrow B}$ is a cofibration) and ${f: A \rightarrow Y}$. Then ${Y \rightarrow Y \cup_f B}$ is a cofibration.

Proof: Again, this can be done via categorical nonsense. In fact, the process of attaching spaces is a special case of the push-out. We can draw a cocartesian diagram

Indeed, this diagram is cocartesian precisely because ${Y \cup_f B}$ is precisely what you get when you combine ${Y}$ and ${B}$ and glue together the things coming from ${A}$. And now it is clear that since ${ A \rightarrow B}$ is a cofibration, ${Y \rightarrow Y \cup_f B}$ is as well. $\Box$

CW pairs

Now, we have to show more generally that a relative CW-complex ${(X, A)}$ is an NDR pair. This will be of crucial importance as we continue discussing homotopy theory, because oftentimes we will construct homotopies piece-by-piece over various skeleta, and we will need to know that they can be extended.

So let’s recall what a relative CW complex is. This is a pair ${(X, A)}$ with a filtration

$\displaystyle A = X^{-1} \subset X^{0} \subset X^1 \subset \dots \subset X$

such that, for each ${k}$, ${X^k}$ is obtained from ${X^{k-1}}$ from attaching ${k}$-cells via maps ${S^{k-1} \rightarrow X^{k-1}}$, and such that ${X}$ is the inductive limit of the ${X^k}$ as a topological space. The last structure means in particular that ${X}$ has the weak topology on the ${X^k}$: a subset of ${X}$ is open if and only if the intersections with each ${X^k}$ are open in ${X^k}$. For instance, the inclusion of a subcomplex of a CW complex satisfies this.

Proposition 4 Let ${(X, A)}$ be a relative CW complex. Then the inclusion$\displaystyle A \rightarrow X$

is a cofibration.

This is the main result of the post, and it will require a bit of preparation. First, let’s recall that in algebraic geometry, after every property (e.g. finite type, quasi-compactness, flatness, etc.) one has a long list of properties. These properties become a standard routine: property X is stable under base-change, preserved under composition, etc., etc. We have not done that for cofibrations. However, there is one such result which we should note.

Lemma 5 Let ${A \stackrel{f}{\rightarrow} B, B \stackrel{g}{\rightarrow} C}$ be cofibrations. Then ${A \stackrel{g \circ f}{\rightarrow} C}$ is a cofibration.

Proof: This is another purely formal statement. Suppose we have ${A \times I \rightarrow X}$ and ${q: C \rightarrow X}$ which are compatible (i.e. agree on ${A}$). Then we have to extend the homotopy ${A \times I \rightarrow X}$ to ${C \times I \rightarrow X}$. To do this, we start by extending it to ${B \times I \rightarrow X}$ (in such a way as to coincide with ${q \circ g}$ on ${B \times \left\{0\right\}}$), then extend this to ${C \times I \rightarrow X}$. So we use the homotopy extension property for each piece twice. $\Box$

Now we are going to use the fact that relative CW complexes are obtained by a repeated attaching procedure. So we show:

Lemma 6 Let ${A}$ be an indexing set. Then the map$\displaystyle \sqcup_A S^{n-1} \hookrightarrow \sqcup_A D^n$

is a cofibration.

Proof: It is easy to check, by the usual routine (and here I shall not spell out the details) that a disjoint union of cofibrations is a cofibration; this comes from the fact that the disjoint union ${\sqcup}$ is a coproduct in the category of topological spaces.

Anyway, this means that it’s enough to show that ${ S^{n-1} \rightarrow D^n}$ is a cofibration. For that, we note that the function ${u(x) = \min (1, 2(1 - ||x||))}$ on ${D^n}$ is such that ${u^{-1}([0, 1))}$ deformation retracts onto ${D^n}$. Moreover, we can arrange things so that the map extends to all of ${D^n}$ (but it is no longer a deformation retraction). In this way, we see that ${(D^n, S^{n-1})}$ is an NDR pair.

Alternatively, we could show that ${S^{n-1} \times I \cup D^n \times \left\{0\right\}}$ is a deformation retract of ${D^n \times I}$. We can do this by “projecting” from a high point about the cylinder ${D^n \times I}$. I should draw a picture, but I am being lazy. $\Box$

Anyway, suppose we have a relative CW complex ${(X, A)}$. Then the previous two propositions together state that for each ${m}$, the map ${A \rightarrow X^m }$ is a cofibration. More generally, if ${m, then ${X^m \rightarrow X^n}$ is a cofibration. This is because it is a composite of attaching maps of disks along their boundaries; we know that composites preserve cofibrations; we know that the inclusion of the sphere into a disk is a cofibration; and we know that attaching leads to cofibrations.

What is now left is to show that ${A \rightarrow X}$ is a cofibration. This will follow from the next proposition. With it, we will have completed the proof that a relative CW complex is a cofibration.

Proposition 7 Suppose ${X}$ is the union of a sequence of subspaces ${X_1 \subset X_2 \subset \dots}$ with the weak topology. If the maps ${X_k \rightarrow X_{k+1}}$ are cofibrations, then ${X_1 \rightarrow X}$ is a cofibration.

Proof: Indeed, let us verify that the pair ${(X, X_1)}$ satisfies the homotopy extension property. Suppose we have a map ${f: X \rightarrow Y}$ and a homotopy ${X_1 \times I \rightarrow Y}$. Then the map ${f}$ is equivalent to a sequence of continuous maps ${f_k: X_k \rightarrow Y}$. Suppose inductively we have constructed ${H_k: X_k \times I \rightarrow Y}$ extending ${H_{k-1}, f_k}$. Then we can use the HEP to extend ${H_k}$ to ${X_{k+1}}$. Repeating this inductively, we get a compatible set of homotopies ${X_k \times I \rightarrow Y}$ which agree with ${f}$ on ${X_k \times \left\{0\right\}}$. Gluing them together (note that ${X}$ has the weak topology) gives the extension of the initial homotopy. $\Box$