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Why does a convex set have trivial fundamental group?

Without using the definition of contractible spaces, could anyone explain for me why that is true?

Adam Chalumeau
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Take a loop $\gamma :[0,1]\to C$ based at $x_0\in C$. Define the homotopy $$H:[0,1]\times [0,1]\to C$$ by the formula $$H(t,s)=sx_0+(1-s)\gamma(t).$$ This is well defined because $C$ is convex and is is a homotopy between any path $\gamma$ and the constant loop $x_0$, so the fundamental group of $C$ is trivial.

Adam Chalumeau
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    Can you look at this question please (if you have time) https://math.stackexchange.com/questions/3286924/the-difference-between-proving-the-existence-of-the-inverse-of-the-fundamental-g?noredirect=1#comment6761304_3286924 ? –  Jul 10 '19 at 11:46