If an \(n\)-dimensional CW complex \(X\) contains a subcomplex \(Y\) homotopy equivalent to \(S^n\), show that the map \(\pi_n(Y) \to \pi_n(X)\) induced by inclusion is injective.
Solution
Let \(i:Y\hookrightarrow X\), where \(X\) is \(n\)-dimensional and \(Y\simeq S^n\).
Assume first \(n\ge2\). Since \(X\) has no \((n+1)\)-cells, an \(n\)-cycle in the cellular chain complex cannot become a boundary after inclusion in \(X\). Thus the inclusion induces an injection \[ i_*:H_n(Y;\mathbb Z)\hookrightarrow H_n(X;\mathbb Z). \] The Hurewicz map \[ \pi_n(Y)\xrightarrow{\cong}H_n(Y;\mathbb Z) \] is an isomorphism because \(Y\simeq S^n\). If \(\alpha\in\pi_n(Y)\) maps to zero in \(\pi_n(X)\), naturality of Hurewicz gives \[ i_*h(\alpha)=h(i_*\alpha)=0. \] Injectivity on \(H_n\) implies \(h(\alpha)=0\), hence \(\alpha=0\). Therefore \(i_*:\pi_n(Y)\to\pi_n(X)\) is injective.
For \(n=1\), \(X\) is a graph, so \(\pi_1(X)\) is free and hence torsionfree. The generator of \(\pi_1(Y)\cong\mathbb Z\) has nonzero image in \(H_1(X)\) by the same cellular-chain argument, so its image in \(\pi_1(X)\) is nontrivial and therefore has infinite order. Thus \(\mathbb Z\to\pi_1(X)\) is injective.
Hence in all cases \[ \boxed{\pi_n(Y)\hookrightarrow\pi_n(X).} \]