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

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- Nov 8th 2009, 03:59 PMTTBHomotopic Functions...
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 - Nov 8th 2009, 07:45 PMaliceinwonderland
Let $\displaystyle f,g:A \rightarrow B$; let $\displaystyle F:A \times I \rightarrow B$ be their homotopy such that $\displaystyle F(a, 0) = f(a)$ and $\displaystyle F(a, 1) = g(a)$ for each a in A. The mapping cylinder $\displaystyle M_f$ for a map $\displaystyle f:A \rightarrow B$ is the quotient space of disjoint union $\displaystyle (A \times I) \coprod B$ obtained by identifying each $\displaystyle (a, 1) \in A \times I$ with $\displaystyle f(a) \in B$. The mapping cylinder $\displaystyle M_f$ deformation retracts to the subspace B by sliding each point (a, t) along the segment $\displaystyle \{a\} \times I \subset M_f$ to the endpoint $\displaystyle f(a) \in B$. We can see easily that B is a deformation retract of both $\displaystyle M_f$ and $\displaystyle M_g$ (We can have $\displaystyle f(A)$ as either F(A,0) or F(A, 1) and g(A) as either F(A,1) or F(A, 0), respectively). Thus $\displaystyle M_f$ and $\displaystyle M_g$ are homotopy equivalent spaces.

- Nov 8th 2009, 07:57 PMTTB
Is it THAT simple? :D 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! :D

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 - Nov 8th 2009, 08:11 PMTTB
...but wait, that doesn't seem to require the condition that f & g are homotopic. (Worried)

- Nov 8th 2009, 08:27 PMaliceinwonderland
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. - Nov 8th 2009, 08:42 PMTTB
Awesome! :D Thank you!