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It is my understanding that, given some filtration $(X^n)_{n \geq 0}$ of a space $X$, or more generally a sequence $X^0 \to X^1 \to X^2 \to \cdots$ of spaces and maps, one usually constructs the "telescope" of the sequence as the subspace $\bigcup_{n \geq 0} [n,n+1] \times X^n \subset [0,\infty) \times X$ (or, in the general case, as a sequence of mapping cylinders glued together at their ends). I am trying to relate this to the following construction in Switzer's Algebraic Topology:

image

I am confused by the presence of smash products and the added disjoint points on the intervals (the notation $A^+$ means $A$ with an added disjoint point), and also the fact that the basepoints are not explicitly mentioned. Is it to be understood that the basepoint of $[n-1,n]^+$ is the added point, or some endpoint of the interval? Either possibility yields a distinct smash product with the $n$-skeleton $X^n$, but in any case I don't see how the end result is supposed to look remotely like the usual telescope. What is the intuition behind this construction? Why smash the $n$-skeleton with an interval with an added disjoint point? Any insight would be greatly appreciated.

Alex Provost
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1 Answers1

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The notation $X^+$ always considers the disjointly added point as the marked point. Switzer's construction is homotopy equivalent to the one that you describe, the difference is in the formal category-theoretical properties. Algebraic topology is generally concerned with pointed spaces: you can't define homotopy groups otherwise, the cohomology are also better behaved. The universal way to convert a space to a pointed space is precisely to add a disjoint base point. Formally $X^+$ is the left adjoint functor to the forgetful functor from pointed spaces to spaces. The telescope can be formally defined as the homotopy colimit of the sequence in question. A homotopy colimit in topology can be explicitly constructed as a usual colimit aka glueing of spaces multiplied by the intervals. In the nonpointed category this reduces to the construction that you described, in the pointed category it reduces to Switzer's one. As a more elemetary motivation, note that you need the telescope to be equipped with a base point and it must be compatible with the base points of $X^i$. The simplest way to achieve that is to collapse all subspaces of the form $[n,n+1]\times x_i,\ x_i \in X^i$ into a single point. This is exactly what Switzer's telescope does.

Anton Fetisov
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  • Thank you for the answer, which is very enlightening. I still have some questions, if you don't mind. I can see why it is desirable to have a universal way of converting any space into a pointed space, but is this generality really needed here? We already have a basepoint $x_0$ singled out in $X$, so isn't ${-1} \times {x_0}$ a natural basepoint to choose in the original telescope if we want to work in the pointed category? Or is it maybe an issue that this basepoint does not belong to the other "building blocks"? – Alex Provost Apr 08 '17 at 02:25
  • @AlexProvost You could certainly do this in principle, since all of these choices are homotopy equivalent, but the canonical ones have better formal properties. You need to define the maps $r_n$, $r$ and $k_n$ and to have them induce an isomorphism on homotopy groups. For $r_n$ and $k_n$ it is not mandatory, since their definition and explicit homotopies don't seem to depend on the basepoints (but I should probably double check that the homotopies are well defined on product construction... which I won't). – Anton Fetisov Apr 08 '17 at 02:43
  • For $r$ you need a commutative square of homotopy groups which is problematic without a good compatible choice of base points. I have a feeling that any construction of this commutative square that would work generally would factor through the smash construction anyway. – Anton Fetisov Apr 08 '17 at 02:43
  • I drew a rough sketch here. Would you mind telling me if my intuition seems correct or if I am making a glaring mistake? Here $X = D^2$, with CW-structure $X^{-1} = {x_0}$, $X^0 = {x_0,x_1}$, $X^1 =$ two arcs connecting the two points in $X^0$, and $X^2 = X$ is the whole disk. The first line depicts the ordinary telescope; the second line depicts the pointed telescope pre-smash product; and the last line depicts the smash product version. (One should imagine that the ends of the adjacent subspaces $[n-1,n]^+ \wedge X^n$ are to be glued together.) – Alex Provost Apr 11 '17 at 20:39
  • @AlexProvost Looks correct to me. – Anton Fetisov Apr 11 '17 at 21:00