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    Isomorphism on $\pi_i$ for $i \leq n$ in $n$-dimensional complexes

    E-HAT-4.2-13

    Area
    Topology
    Topics
    Higher Homotopy Groups
    Status
    draft

    Show that a map between connected \(n\)-dimensional CW complexes is a homotopy equivalence if it induces an isomorphism on \(\pi_i\) for \(i \leq n\).

    Solution

    Let \[ f:X\to Y \] be a map of connected \(n\)-dimensional CW complexes inducing isomorphisms on \(\pi_i\) for \(i\le n\). In particular it is an isomorphism on \(\pi_1\), so it lifts to a map of universal covers \[ \widetilde f:\widetilde X\to\widetilde Y. \] The universal covers are simply connected \(n\)-dimensional CW complexes, and \(\widetilde f\) is an isomorphism on \(\pi_i\) for \(2\le i\le n\).

    Consider its mapping-cylinder pair \((M_{\widetilde f},\widetilde X)\). The long exact sequence of homotopy groups shows that this pair is \(n\)-connected. Relative Hurewicz therefore gives \[ H_i(M_{\widetilde f},\widetilde X)=0 \qquad(i\le n). \] Equivalently, \[ \widetilde f_*:H_i(\widetilde X)\xrightarrow{\cong}H_i(\widetilde Y) \qquad(i\le n). \] For \(i>n\), both homology groups vanish because the two universal covers are \(n\)-dimensional CW complexes. Hence \(\widetilde f\) is a homology isomorphism in all degrees.

    Exercise 12 now applies: since \(f\) is an isomorphism on \(\pi_1\) and its lift is a homology equivalence, \[ \boxed{f\text{ is a homotopy equivalence}.} \]