Cone in topological space Homotopy problem

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SUMMARY

The discussion focuses on the homotopy properties of the cone on a topological space Y, denoted as CY. It establishes that any two continuous functions f and g from a space X to CY are homotopic, which is a crucial result in algebraic topology. Additionally, part (b) of the problem seeks to find the fundamental group (π1) of CY at a point p, which is directly derived from the conclusions of part (a). The participants emphasize the importance of identifying a special function k: X → CY to facilitate the homotopy between f and g.

PREREQUISITES
  • Understanding of topological spaces and continuous functions
  • Familiarity with homotopy theory in algebraic topology
  • Knowledge of fundamental groups, specifically π1
  • Experience with cone constructions in topology
NEXT STEPS
  • Study the properties of homotopy equivalence in topological spaces
  • Learn about the construction and properties of cones in topology
  • Explore the concept of fundamental groups and their applications
  • Investigate examples of homotopies between continuous functions
USEFUL FOR

Mathematics students, particularly those studying algebraic topology, topologists, and anyone interested in the properties of continuous functions and homotopy theory.

JoeSabs
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Homework Statement



Let Y be a topological space. Let CY denote the cone on Y.

(a) Show that any 2 continuous functions f, g : X --> CY are homotopic.
(b) Find (pi)1 (CY, p).

Homework Equations



I have no idea. The professor said one problem would be way out in left, to see who could make the connections. I can't. lol

The Attempt at a Solution



See 2.!

I know I haven't made an attempt, so I'm not asking for an answer. Any hints or help is much appreciated.
 
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Remember that if two functions f, g are each homotopic to another function k, then you can compose the homotopies to see that f is homotopic to g. Keeping that in mind, can you find a function k: X \to CY which is special somehow, and useful for making homotopies in this way?

Part (b) is an easy corollary of part (a).
 

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