Hi,
Suppose f & g are both functions mapping from A to B. (A & B are topological spaces.)
Also assume that f & g are homotopic.
Then how does one prove that the Mapping Cylinders of f & g are homotopic spaces?
Thanks x
Let ; let be their homotopy such that and for each a in A. The mapping cylinder for a map is the quotient space of disjoint union obtained by identifying each with . The mapping cylinder deformation retracts to the subspace B by sliding each point (a, t) along the segment to the endpoint . We can see easily that B is a deformation retract of both and (We can have as either F(A,0) or F(A, 1) and g(A) as either F(A,1) or F(A, 0), respectively). Thus and are homotopy equivalent spaces.
Is it THAT simple? Y is a deformation retract of both Mf & Mg, so Y is homotopic to Mf & Mg & therefore, as homotopy is an equivalence relation, Mf is homotopic to Mg?
Cool!
Thank you!
Are you sure there's nothing else to it? I spent aaaaaaaaaaaaages playing around with homtopies like F:Mfx[0,1] ---> Mg, etc. Do you not need to do ANY of that?? x
By convention, we say a map f is "homotopic" to a map g and a topological space A is "homotopy equivalent to" a topological space B. Rather than saying "Y is homotopic to M_f", it is more appropriate to say "Y is homotopy equivalent to M_f" (Take M_f as a quotient topological space as mentioned in the previous post).
I think that if f and g are topological continuous mappings from A to B, then both f(A)=g(A)=B are ensured and we don't need a homotopy condition between f and g. Meanwhile, if there is a homotopy condition between f and g, then topological continuity of f and g are ensured, so f(A)=g(A)=B.