# Homotopic Functions...

• November 8th 2009, 04:59 PM
TTB
Homotopic 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
• November 8th 2009, 08:45 PM
aliceinwonderland
Quote:

Originally Posted by TTB
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 $f,g:A \rightarrow B$; let $F:A \times I \rightarrow B$ be their homotopy such that $F(a, 0) = f(a)$ and $F(a, 1) = g(a)$ for each a in A. The mapping cylinder $M_f$ for a map $f:A \rightarrow B$ is the quotient space of disjoint union $(A \times I) \coprod B$ obtained by identifying each $(a, 1) \in A \times I$ with $f(a) \in B$. The mapping cylinder $M_f$ deformation retracts to the subspace B by sliding each point (a, t) along the segment $\{a\} \times I \subset M_f$ to the endpoint $f(a) \in B$. We can see easily that B is a deformation retract of both $M_f$ and $M_g$ (We can have $f(A)$ as either F(A,0) or F(A, 1) and g(A) as either F(A,1) or F(A, 0), respectively). Thus $M_f$ and $M_g$ are homotopy equivalent spaces.
• November 8th 2009, 08:57 PM
TTB
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
• November 8th 2009, 09:11 PM
TTB
...but wait, that doesn't seem to require the condition that f & g are homotopic. (Worried)
• November 8th 2009, 09:27 PM
aliceinwonderland
Quote:

Originally Posted by TTB
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

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).

Quote:

Originally Posted by TTB
...but wait, that doesn't seem to require the condition that f & g are homotopic. (Worried)

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.
• November 8th 2009, 09:42 PM
TTB
Awesome! :D Thank you!