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Feb 22, 2017 at 21:33 comment added HJRW @JensReinhold, since your comment seems to answer the question, why don't you post it as an answer?
Feb 22, 2017 at 8:59 history edited Oscar Randal-Williams CC BY-SA 3.0
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Feb 21, 2017 at 20:59 comment added Oscar Randal-Williams I suppose that is right. I had mistakenly thought that because G is simply-connected that 2-cells and higher did not matter for the fundamental group , but of course the trivial group shows this is not the case...
Feb 21, 2017 at 20:19 comment added Jens Reinhold I am skeptical about this argument. Why can we assume that there are only G-cells of dimension $\leq 1$? Actually, why can't we start with $G/(\mathbb Z/k)$ for $k$ arbitrarily large (that is the unique 0-cell) and attach a 2-cell $G \times D^2$ via a map that equivariantly extends a generator $S^1 \to G/(\mathbb Z/k)$ of the fundamental group $\pi_1(G/(\mathbb Z/k)) = \mathbb Z/k$? This example should answer my question in the negative.
Feb 21, 2017 at 16:55 vote accept Jens Reinhold
Feb 21, 2017 at 17:32
Feb 20, 2017 at 20:10 history answered Oscar Randal-Williams CC BY-SA 3.0