Eeeeaaargh (homotopy, what else)

[quasar987]

Homework Statement

Question in my 3 hours final from 10 minutes ago that I thought about for 2 hours:

Let f,g,h:X-->X be continuous fonctions. If f o h ~ g o h, is it true that f ~ g.

The Attempt at a Solution

Nothing convincing.

 

[HallsofIvy]

Of course not. Take h to be a "constant" function: h(x)= p for all x. Then
f o h= g o h for all f and h.

 

[Hurkyl]

Homework Statement

Question in my 3 hours final from 10 minutes ago that I thought about for 2 hours:

Let f,g,h:X-->X be continuous fonctions. If f o h ~ g o h, is it true that f ~ g.

The Attempt at a Solution

Nothing convincing.

Observe that what you just wrote is the definition of "h is a epimorphism". (in the category of spaces and maps modulo homotopy)

 

[AKG]

What does the ~ stand for?

 

[matt grime]

Is homotopic to (as the title of the post implies).

 

[quasar987]

Of course not. Take h to be a "constant" function: h(x)= p for all x. Then
f o h= g o h for all f and h.

You meant "for all f and g"?

And how is this possible. Let f(p)=q and g(p)=r.

And finally, even if you're right, what does this prove?

 

[StatusX]

But what does this prove?

He meant f and g, and this means f and g need not have any relation to each other (other than that f(p) can be joined by a path to g(p)) so if you can find any two that aren't homotopic, you'll have a counterexample.

This is the same kind of confusion in the other thread, namely, if you are wondering whether A=>B, but you see you can get A in an almost completely generic situation, then it almost always follows that this implication is false, you just need to find one example that satisfies B and one that doesn't (and which both satisfy A, which wll follow from the genericness).

 

[matt grime]

What do you mean what does it prove? It is a counter example. Since fh=gh for all f and g, you certainly can't cancel on the right, unless you think all maps are homotopic.

 

[quasar987]

In my exam, I said

Let X=[0,1]u[2,3], let h be the constant function h(x)=0, let f be the identity and g(x)=x if x is in [0,1] and g(x)=x-2 if x is in [2,3].

Then (f o h)(x) = (g o h)(x) = 0 for all x, so f o h ~ g o h. And then I said, not half convinced, that f is not ~ to g.

Is that any good? It kind of ressembles what Halls of Ivy said, except he said (f o h)(x) = (g o h)(x) no matter f and g which I can't make any sense of even if my life depended on it.

 

[matt grime]

Actually, I mispoke. If h is constant, the fh is constant, and gh is constant for any f,g. Let's suppose these are homotopic, i.e. the points lie in the same path component, and fh~gh. Then are you saying that you think *all* maps must be homotopic? Clearly this can't be: let f be the identity map and g be *any* map not homtopic to the identity. Surely you can think of spaces that are pathconnected (so all constant maps are homotopic), and that have two non-homotopic maps.

 

[AKG]

In my exam, I said

Let X=[0,1]u[2,3], let h be the constant function h(x)=0, let f be the identity and g(x)=x if x is in [0,1] and g(x)=x-2 if x is in [2,3].

Then (f o h)(x) = (g o h)(x) = 0 for all x, so f o h ~ g o h. And then I said, not half convinced, that f is not ~ to g.

Is that any good?

Yeah, that sounds good.

It kind of ressembles what Halls of Ivy said, except he said (f o h)(x) = (g o h)(x) no matter f and g which I can't make any sense of even if my life depended on it.

What he should have said is this (which is exactly what matt grime also just said):

Let h be constant, with value c. If f(c) and g(c) lie in the same path-connected component of X, then f o h ~ g o h. This holds for ANY two functions f and g such that f(c) and g(c) lie in the same p-c component of X. This is a very weak condition to impose on f and g, so chances are that there are many function f and g which aren't homotopic to one another, even though f(c) and g(c) can be path-connected.