I'm reading Ravenel's book Nilpotence and periodicity in stable homotopy theory. In section 1.1, it says the set of homotopy classes of maps of maps between compact manifolds or between algebraic varieties over real or complex numbers is countable. I know that the set of homotopy classes between triangulable spaces is countable. I also know that algebraic varieties over real or complex numbers are triangulable while compact manifolds are not always triangulable. So I want to know how to prove that the set of homotopy classes of maps between compact manifolds is countable.
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$\begingroup$ You don't need triangulabilty, but merely that the homotopy class is determined by a finite set in the source and a countable set in the target. $\endgroup$– Igor RivinCommented Dec 16, 2016 at 2:12
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9$\begingroup$ Smooth manifolds are triangulable. Topological manifolds are homotopy equivalent to CW complexes. Or a simpler way to treat topological manifolds for this purpose might be to observe that they are ENRs and therefore are retracts of triangulable spaces. $\endgroup$– Tom GoodwillieCommented Dec 16, 2016 at 2:33
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4$\begingroup$ To expand on Tom Goodwillie's comment, the key point is that functoriality of $[X,Y]$, the set of homotopy classes of maps $X\to Y$, implies that if $X$ is a retract of $X'$ and $Y$ is a retract of $Y'$, then $[X,Y]$ injects into $[X',Y']$. Since compact manifolds are retracts of finite simplicial complexes (as shown in Corollaries A.8 and A.9 of my book for example) this reduces countability of $[X,Y]$ when $X$ and $Y$ are compact manifolds to countability when they are finite simplicial complexes. $\endgroup$– Allen HatcherCommented Dec 16, 2016 at 19:06
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$\begingroup$ @AllenHatcher: Thank you for your answer and your book from which I have learned a lot. $\endgroup$– BorromeanCommented Dec 17, 2016 at 10:45
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This may be somewhat heavy-handed but it follows from work of Kirby-Siebenmann that any compact manifold is homotopy equivalent to a finite CW-complex (implying what you want because any finite CW-complex is homotopy equivalent to its regular neighborhood in some Euclidean space, which in turn can be triangulated as a finite simplicital complex).
More precisely, let $M$ be a compact manifold, possibly with boundary. If $M$ has dimension $\ge 6$ and empty boundary, then $M$ is homeomorphic to a CW-complex [p.107 of Kirby-Siebenmann's book "Foundations of Topological manifolds"]. In general, fix $k$ such that $W=M\times D^k$ has dimension $\ge 7$. On the same page 107 it is stated that $W$ is homeomorphic to a mapping cylinder of a map $f:\partial W\to X$, where $X$ is a finite CW-complex. As mentioned above $\partial W$ is also a CW complex. The map $f$ can be homotoped to a cellular map, so the mapping cylinder $W$ is homotopy equivalent to a CW-complex as promised.
answered Dec 16, 2016 at 2:15
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$\begingroup$ Thanks a lot. I have already found these results in Hatcher's algebraic topology. Theorem 2C.5 and Corollary A.12 together answer my question. $\endgroup$ Commented Dec 16, 2016 at 6:27
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2$\begingroup$ @ZheyanWan: How does it answer the question? Note that A.12 does not claim that any compact manifold is homotopy equivalent to a finite CW complex. It just claims homotopy equivalence to some CW complex. (Naively one might think that this CW complex surely can be chosen finite. What is true and easy to show is that any compact manifold is dominated by a finite complex, but the latter property is not the same as being homotopy equivalent to a finite complex because of Wall's finiteness obstruction). $\endgroup$ Commented Dec 16, 2016 at 15:11
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$\begingroup$ Thank you for your answer and comment. This is a good starting point for me to study algebraic K-theory. $\endgroup$ Commented Dec 17, 2016 at 10:48