Let $f, g$ be two paths in a space such that their concatenation $f * g$ is nullhomotopic. Prove that $f$ is homotopic to $g$ rel $\{0,1\}$.
Homotopy of two paths whose composition is nullhomotopic
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algebraic-topology
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0You can't prove it, because it's not true. – 2010-12-14
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1To elabore on Ryan's comment: Your statement becomes ture when you have to replace $g$ with its reversed path. – 2010-12-14
1 Answers
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As Rasmus points out, once you state the result correctly, it's easy -assuming you've already proved some elementary properties of homotopy.
Namely, let $f,g$ be two loops with $f(0) = g(0) = f(1)= g(1) = x_0$ and such that $g^{-1}*f \sim 1$, where $1$ is the constant path at $x_0$.
Then $f\sim 1* f \sim (g*g^{-1})*f \sim g* (g^{-1}*f) \sim g*1 \sim g$.
Exercise. Which "elementary properties of homotopy" have we used and where?