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I think I'm getting burnt out, as I'm having trouble understanding Spanier's definition of a relative CW complex (from his Algebraic Topology): A relative CW complex $\big(X, A \big)$ consists of a topological space $X$, a closed subspace $A$ and a sequence of closed subspaces $\big( X, A \big)^k$ for $k \ge 0$ such that:


(1) $(X, A)^0$ is obtained from $A$ by adjoining 0-cells.

(2) For $k \ge 1$, $\big( X, A \big)^k$ is obtained from $\big( X, A \big)^{k-1}$ by adjoining k-cells.

(3) $X = \bigcup (X,A)^k$.

(4) $X$ has a topology coherent with $\{ \big( X, A \big)^k \}_k$


For (1, 2) , how does this contrast with the usual definition of constructing a CW complex (The sort of conventional definition you might see in, say, Hatcher.)? What does it mean to take a union across this ordered pair, when its meant to resemble two distinct topological spaces?

For (4) I know what the coherent topology is, but I don't know what Spanier wants to represent by writing a $_k$ in $\{ \big( X, A \big)^k \}_k$


As a bit of a technical background, I'm an engineer who is ultimately trying to get back to dealing with the covariant derivative on manifolds. I need an intuitive (visualizable) understanding of what this construction is for each element. For CW complexes I can obviously just imagine taking a dot, taping a circle to it, then taping a disk minus its boundary to make 'a' surface. Ditto for a torus, except you tape two circles to the point. Can we construct a surface (maybe a torus or something simple and similar) that helps illustrate this example for $X$ and $A$? I need something to "click around" in my head like the torus and surface example more than I need any kind of proof. Thanks in advance.

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  • $\begingroup$ Well, you could take $A = S^1 \vee S^1$ and then attach a 2-cell to make a torus. It's the same as the construction you describe, except the 1-skeleton is now simply assumed given. The notion is generally more useful when you do not know whether $A$ has a CW-structure or not (or even know for sure that it doesn't), for instance when constructing a Whitehead tower for $A$. $\endgroup$ Feb 10 at 16:45

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First let me correct your intuition of a CW complex, and then I'll explain how to extend the corrected intuition to a relative CW complex.

For a CW complex $X$ you start from nothing.

Next, you add not just one dot but a whole bunch of dots. That's the $0$-skeleton $X^0$.

Next, you glue on a bunch of line segments (not circles); each endpoint of each line segment is glued to a dot, and the two endpoints of a line segment do not have to be glued to the same dot. That's the $1$-skeleton $X^1$.

Next, you tape on some disks: the boundary circle of each disc is taped to the $1$-skeleton. That's the $2$-skeleton $X^2$.

And so on.

In the end you get the topological space $X$ and a sequence of subspaces denoted $$\emptyset \subset X^0 \subset X^1 \subset X^2 \subset \cdots $$ and the union of this sequence of subsets is $X$ itself: $$X = \emptyset \cup X^0 \cup X^1 \cup X^2 \cup \cdots $$ which I can write more succinctly as $$X = \bigcup_{k \ge 0} X^k $$ And why did I bother throwing the emptyset into that description? It would have been simpler to just leave it out, and nothing would have changed. Well....... just wait.


For a relative CW complex, the definition is every so slightly amended: instead of starting from nothing, you start from some topological space $A$. Think of $A$ as just some blob. Also, think of an ordinary CW complex as the special case where the blob $A$ is the emptyset.

As with an ordinary CW complex, the definition of a relative CW complex proceeds by describing how the $k$-skeleton is constructed from the $k-1$ skeleton. That $k$-skeleton is denoted $(X,A)^k$. But let me warn you not to over-interpret the notation $(X,A)^k$: it stands for a certain subset of $X$; the union of this sequence of subsets equals the whole of $X$, just as in the definition of an ordinary CW complex.

Okay, so here we go.

For a relative CW complex $(X,A)$ you start from the topological space $A$.

Next, you add to $A$ not just one dot but a whole bunch of dots. That's the $0$-skeleton $(X,A)^0$.

Next, you glue on a bunch of line segments. Each endpoint of each line segment is glued to a point of $(X,A)^0$. The point of $(X,A)^0$ that endpoint is glued to can be one of the dots that you added to $A$, or can be a point of $A$ itself. That's the $1$-skeleton $(X,A)^1$.

Next, you tape on some discs. The boundary of each circle is taped to the $1$-skeleton $(X,A)^1$. That's the $2$-skeleton $(X,A)^2$.

And so on.

In the end you get the topological space $X$ and a sequence of subspaces denoted $$A \subset (X,A)^0 \subset (X,A)^1 \subset (X,A)^2 \subset \cdots $$ and the union of this sequence of subsets is $X$ itself: $$X = A \cup (X,A)^0 \cup (X,A)^2 \cup (X,A)^2 \cup \cdots $$ I could also have written this as $$X = A \cup \bigcup_{k \ge 0} (X,A)^k $$ although since $A \subset (X,A)^0$ it's not changing anything to write this as $$X = \bigcup_{k \ge 0} (X,A)^k $$


Finally, the notation $\{(X,A)^k\}_k$ is just a shorthand for "set builder" notation which might be written out more fully as $$\{(X,A)^k \mid k=0,1,2,3,...\} \quad\text{or as}\quad \{(X,A)^k \mid k \ge 0\} $$ or something like that. In other words, that notation simply stands for the set of skeleta of the relative CW complex $(X,A)$.

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  • $\begingroup$ Fantastic answer Lee, I think I understood everything. Thank you! A slightly pedantic question: You are using $\subset$ in the old school way in which it doesn't necessarily mean a proper subset, right: $\big( \text{"} \subset \text{"} \; \dot = \; \text{"} \subseteq \text{"} \big)$. I'm asking that because for the CW complex, I could have one point in $X^0$, NO lines in $X^1$, and one surface in $X^2$, which makes a sphere $\mathbb S^2$ but has a sequence $X^0 = X^1 \subset X^2$ or $X^0 \subseteq X^1 \subseteq X^2$. $\endgroup$
    – Nate
    Feb 11 at 1:15
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    $\begingroup$ Yes, that's right. In general, for any ordinar CW complex $X$, and for any $k \ge 0$, you can think of $X$ as a CW complex relative to $A=X^k$ such that there are no cells of any dimension between $0$ and $k$. $\endgroup$
    – Lee Mosher
    Feb 11 at 2:30

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