Let \(X\) be a connected CW complex with 1-skeleton \(X^1\). Show that \(\pi_2(X, X^1) \approx \pi_2(X) \times K\) where \(K\) is the kernel of \(\pi_1(X^1) \to \pi_1(X)\), a free group. Show also that the map \(\pi_2'(X) \to \pi_2'(X, X^1)\) need not be injective by considering the case \(X = \mathbb{RP}^2\) with its standard CW structure.
Solution
Since \(X^1\) is a graph, \[ \pi_2(X^1)=0. \] The long exact sequence of the pair gives a short exact sequence \[ 0\longrightarrow\pi_2(X) \longrightarrow\pi_2(X,X^1) \xrightarrow{\partial}K\longrightarrow1, \] where \[ K=\ker\bigl(\pi_1(X^1)\to\pi_1(X)\bigr). \] The group \(K\) is a subgroup of the free group \(\pi_1(X^1)\), hence is free by Nielsen–Schreier. Choose a free basis of \(K\) and choose one lift in \(\pi_2(X,X^1)\) of each basis element. By the universal property of a free group these lifts extend to a homomorphic section \[ s:K\to\pi_2(X,X^1) \] of \(\partial\).
By Exercise 27, the image of \(\pi_2(X)\) in \(\pi_2(X,X^1)\) is central. Hence the resulting semidirect product is in fact direct: \[ \boxed{\pi_2(X,X^1)\cong\pi_2(X)\times K.} \]
For the second assertion take \(X=\mathbb{RP}^2\) with its standard CW structure and \(X^1=S^1\). The action of \[ \pi_1(X)\cong\mathbb Z/2 \] on \[ \pi_2(X)\cong\pi_2(S^2)\cong\mathbb Z \] is multiplication by \(-1\). Therefore \[ \pi_2'(X)\cong\mathbb Z/(a\sim-a)\cong\mathbb Z/2. \] On the other hand the pair \((X,X^1)\) is obtained by attaching one \(2\)-cell, so the relative Hurewicz theorem gives \[ \pi_2'(X,X^1)\cong H_2(X,X^1)\cong\mathbb Z. \] The induced homomorphism \[ \mathbb Z/2\longrightarrow\mathbb Z \] must be zero, hence is not injective. This is the required counterexample.