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    Factoring maps through $n$-connective covers

    E-HAT-4.1-16

    Area
    Topology
    Topics
    Higher Homotopy Groups
    Status
    draft

    Show that a map \(f: X \to Y\) between connected CW complexes factors as a composition \(X \to Z_n \to Y\) where the first map induces isomorphisms on \(\pi_i\) for \(i \leq n\) and the second map induces isomorphisms on \(\pi_i\) for \(i \geq n+1\).

    Solution

    Use the relative cell-attachment construction from the proof preceding Proposition 4.13. Starting with \[ f:X\longrightarrow Y, \] attach cells to the source only in dimensions at least \(n+1\), extending the map over every attached cell.

    Inductively, suppose a map \[ f_k:Z_k\to Y \] has already been made an isomorphism on \(\pi_i\) for \[ n+1\le i<k. \] Attach \(k\)-cells corresponding to generators of the relative group \[ \pi_k(Y,Z_k) \] so that the resulting map is surjective on \(\pi_k\). Then attach \((k+1)\)-cells along representatives of the kernel of \[ \pi_k(Z_k)\to\pi_k(Y) \] and extend them over \(Y\) by chosen nullhomotopies. The resulting map is an isomorphism on \(\pi_k\), while the already established lower homotopy groups are unchanged.

    Begin this process at \(k=n+1\) and continue for all \(k\). Let \(Z_n\) be the union. Since every cell added to \(X\) has dimension at least \(n+1\), cellular approximation for the pair \((Z_n,X)\) gives \[ \pi_i(Z_n,X)=0\qquad(i\le n), \] so the long exact sequence of the pair yields \[ \pi_i(X)\xrightarrow{\cong}\pi_i(Z_n) \qquad(i\le n). \] By the inductive construction, the extension \[ g:Z_n\to Y \] induces isomorphisms \[ \pi_i(Z_n)\xrightarrow{\cong}\pi_i(Y) \qquad(i\ge n+1). \] The original map is the composite \[ X\longrightarrow Z_n\xrightarrow{g}Y. \] Thus \[ \boxed{f\text{ has the required factorization}.} \]