What is the fundamental group of a "kite-shaped" two dimensional figure, with lines connecting opposite pairs of corner-points? (So, a diamond with a cross in the middle.)
Doesn't this just deformation retract to a wedge of 4 circles?
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asked Sep 12, 2014 at 20:28
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$\begingroup$ Yes, it does. Alternatively, you can calculate the Euler characteristic and use that to compute $H_1$ and hence $\pi_1$. This method is useful for more complicated graphs. $\endgroup$– Ayman HouriehSep 12, 2014 at 20:32
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$\begingroup$ Ok. This was on my qualifier exam and I wanted to make sure I got the correct answer. Thanks! $\endgroup$– Johnny AppleSep 12, 2014 at 20:33
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1 Answer
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Yes, it deformation retracts to a wedge of four circles. Additionally, $\pi_1(\bigvee^4 S^1)=\ast^4\pi_1(S^1)$, where $\ast$ is the free product of groups.