Let G be a topological group and $f,g : [0,1] \times [0,1] \to G$ two loops based at 1. Why isn't the application $$ H:\begin{array}{rcl} I \times I & \longrightarrow & G \\ (t,u) & \mapsto & f(tu)g((1-t)u) \end{array} $$ an homotopy?
2026-04-01 02:01:49.1775008909
In a topological group all loops based at 1 are homotopic
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