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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