Is there a continuous path in function space from f to g?

[madness]

Any two continuous maps from X to Y, where Y is a convex subset of R^n, are homotopic. For example, the functions f(t) = (sin2pit, cos2pit) and g(t) = (t,0) are maps from [0,1) to R^2. So these functions are homotopic. Intuitively, two functions are homotopic if one can be continuously deformed into the other. So is there really a continuous path in function space from f to g?

 

[JSuarez]

As your circle is not closed, because it's the image of [0,1), it can be obtained by stretching and bending the image of g(t), which coincides with the interval [0,1) on the x-axis (the analytic expression of the homotopy is messy).

 

[madness]

Ok how about I define the circle on [0,1] then? The analytic expression of the homotopy is h(t,s) = (1-s)f(t) + sg(t), which is not messy at all.

 

[wofsy]

Any two continuous maps from X to Y, where Y is a convex subset of R^n, are homotopic. For example, the functions f(t) = (sin2pit, cos2pit) and g(t) = (t,0) are maps from [0,1) to R^2. So these functions are homotopic. Intuitively, two functions are homotopic if one can be continuously deformed into the other. So is there really a continuous path in function space from f to g?

how about F(s,t) = sf(t) + (1-s)g(t) ?

 

[madness]

Yes, I assume you didn't see my last post which stated exactly that. My problem is that continuous transformations don't generally involve breaking or tearing, which makes this counter-intuitive. For example, the circle is not homeomorphic to the line, but in terms of functions, it is homotopic to the line.

 

[wofsy]

Yes, I assume you didn't see my last post which stated exactly that. My problem is that continuous transformations don't generally involve breaking or tearing, which makes this counter-intuitive. For example, the circle is not homeomorphic to the line, but in terms of functions, it is homotopic to the line.

sorry I didn't see your post. The homotopy works and just takes advantage of the convexity of the plane.

There is no tearing or breaking here. breaking would happen if for some value of s, sf(t) + (1-s)g(t) broke the domain interval (0,1]. But each image is unbroken.

 

[madness]

Not sure if I follow you there. I was thinking the path in function space should be unbroken, ie the is a continuum of functions from the circle to the line. I don't see what this has to do with the domain.

 

[wofsy]

Not sure if I follow you there. I was thinking the path in function space should be unbroken, ie the is a continuum of functions from the circle to the line. I don't see what this has to do with the domain.

breaking means breaking the domain not the image.You imagine the domain as deformed into the range by the function. If this deformation does not break the domain then the map is continuous

The path in function space is also unbroken. The image of the interval from zero to 1 in function space is unbroken. realize though that you need a topology on the function space to talk about continuous paths of functions - but the convex homotopy will work for reasonable topologies.

What you may be thinking of is a slightly different problem and are getting tripped up because the image of your function,f, is a circle in the plane.

If your domain is a circle then f is a map of this circle into the plane and its image can not be deformed onto the line segment without breaking .But there is no way to extend g to the circle continuously which again shows you that no such homotopy could ever exits on the circle. So it is all about what the function does to the domain.