Hello, I have a question regular values and smooth homotopies. Usually in giving the definition of regular value, they disregard the regular values whose inverse image is empty set (although they should be called regular values if we want to be able to say that set of regular values is dense for smooth functions . Also given a map f: M [itex]\rightarrow [/itex]N a map from some m-dimensional manifold to n-dimensional manifold, and y a regular value [itex]f^{-1}(y)[/itex] is said to be m-n dimensional submanifold and when stating this, they also disregard regular values whose inverse image is the empty set. Now this seems to create a problem, or maybe I am wrong.(adsbygoogle = window.adsbygoogle || []).push({});

Consider the example in milnor's book where he says a orientation reversing diffeomorphism of a compact boundaryless manifold can be smoothly homotopic to the identity (because homotopic maps have same index).

As M and N I take the unit disk [itex]D^2[/itex]. Consider the "homotopy" F: [0,1] x [itex]D^2[/itex] [itex]\rightarrow[/itex] [itex]D^2[/itex]

F(x,y,t) = (x, -1(1-t)y +ty). F(x,y,0) = (x,-y) and F(x,y,1) = (x,y) . First of all why is this not a smooth homotopy. Yes as t changes, range becomes a subset of unit disk and infact at t=1/2 it is the intersection of the unit disk with the y=0 line but nothing is said about the range of maps in smooth homotopy.

Moreover, DF = [1 0 0 ; 0 2t-1 2y] so that for instance z=(0,-1) is a regular value whose inverse image is the two points (0, 1,0) and (0,-1,0). But since z is a regular value shouldn't this be a 1-dimensional submanifold? I think the problem here arises because intersection of [itex]f^{-1}(z)[/itex] with the sets t x [itex]D^2[/itex] is empty for t not equal to 0 or 1.

Thanks

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# Smooth Homotopy, Regular Values (Milnor)

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