The stages of a homotopy are continuous

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Given a homotopy, [ilmath]F:X\times I\rightarrow Y[/ilmath] (for topological spaces [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath], where [ilmath]I[/ilmath] denotes the unit interval, [ilmath]I:=[0,1]\subset\mathbb{R}[/ilmath]), the stages of the homotopy [ilmath]F[/ilmath] are continuous[1]. More concisely:

  • For a homotopy, [ilmath]F:X\times I\rightarrow Y[/ilmath], for all [ilmath]t\in I[/ilmath] the map [ilmath]f_t:X\rightarrow Y[/ilmath] given by [ilmath]f_t:x\mapsto F(x,t)[/ilmath] is continuous.

Overview of proof


  1. Alec's own work


TODO: Mark this as a "low hanging fruit" page - I've outlined the proof below.

[ilmath]\xymatrix{ & X\times I \ar[r]^F & Y \\ X\times\{t\} \ar@{^{(}->}[ur]^\iota \ar@{->}[dr]_{\pi}& & \\ & X \ar[uur]_{f_t} \ar@{.>}@/_/[ul]_-{\pi^{-1}:x\mapsto (x,t)} & }[/ilmath]

This diagram commutes, composition of continuous maps is continuous, thus [ilmath]f_t[/ilmath] is continuous; job done.