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Oct 20, 2020 at 0:04 vote accept Hussain Kadhem
Oct 19, 2020 at 22:12 answer added Joshua Mundinger timeline score: 4
Oct 19, 2020 at 22:11 history edited Hussain Kadhem CC BY-SA 4.0
clarification
Oct 19, 2020 at 22:08 comment added Fernando Muro 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.
Oct 19, 2020 at 21:45 comment added Hussain Kadhem Right, but I don't follow how the inclusion of $S^2$ induces an inclusion of $\mathbb{Z}[G]$. (Also fixed the markup thanks.)
Oct 19, 2020 at 21:44 history edited Hussain Kadhem CC BY-SA 4.0
deleted 2 characters in body
Oct 19, 2020 at 21:41 review Close votes
Oct 24, 2020 at 3:06
Oct 19, 2020 at 21:26 history edited YCor CC BY-SA 4.0
formatting, added tags
Oct 19, 2020 at 21:21 comment added Fernando Muro 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).
Oct 19, 2020 at 21:17 history asked Hussain Kadhem CC BY-SA 4.0