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I am reading a paper which makes the following claim:

let $G$ be a finitely presented group, and let $X$ be the presentation complex of $G$. Let $X' = X \vee S^2$ be the wedge sum of $X$ with the sphere. Then $X'$ induces a map $\alpha: S^2 \to X'$ which generates a free copy of $\mathbb{Z}[G] = \mathbb{Z}[\pi_1(X')]$ inside $\pi_2(X')$. Here, $\mathbb{Z}[G]$ denotes the group ring of $G$ over the integers.

I don't follow with how $\alpha$ includes $\mathbb{Z}[G]$ in $\pi_2(X')$. What is the justification here?

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    $\begingroup$ The map $\alpha$ is just the inclusion of $S^2$ in $X\vee S^2$ (note that the topologist's wedge symbol $\vee$ is \vee, not \wedge). $\endgroup$
    – Fernando Muro
    Commented Oct 19, 2020 at 21:21
  • $\begingroup$ Right, but I don't follow how the inclusion of $S^2$ induces an inclusion of $\mathbb{Z}[G]$. (Also fixed the markup thanks.) $\endgroup$
    – Hussain Kadhem
    Commented Oct 19, 2020 at 21:45
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    $\begingroup$ Hussain, could you then rephrase your question to ask that? A geometric answer would be that the covering space of $X'$ can be obtained from that of $X$ by attaching a copy of $S^2$ to each vertex. $\endgroup$
    – Fernando Muro
    Commented Oct 19, 2020 at 22:08

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Let $\tilde Y$ denote the universal cover of $Y$. Then $\tilde{X'}$ is formed from $\tilde X$ by gluing a copy of $S^2$ to each preimage of the basepoint of $X$. Then by the Hurewicz isomorphism, we have $$ \pi_2(X') = \pi_2(\tilde{X'}) = H_2(\tilde X) \oplus \bigoplus_{g \in \pi_1(X)} \mathbb Z = \pi_2(X) \oplus \mathbb Z[\pi_1(X)],$$ since the classes of the glued spheres are independent and generate a summand of $H_2(\tilde X)$, and are freely permuted by $\pi_1(X)$.

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  • $\begingroup$ Very clear, thanks for the answer! $\endgroup$
    – Hussain Kadhem
    Commented Oct 20, 2020 at 0:04

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