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

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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.

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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

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