Re: Continuity of inverse of continuous function
- From: quasi <quasi@xxxxxxxx>
- Date: Mon, 29 Aug 2005 10:59:22 -0700
On Mon, 29 Aug 2005 07:55:19 -0500, mstemper@xxxxxxxxxxxxxxxx (Michael
Stemper) wrote:
>A fact that I thought that I remembered from Calculus was that if a
>continuous function has an inverse, the inverse is also continuous.
>Looking through Fadell & Fadell doesn't show me such a result, so
>I'm trying to prove it on my own.
>
>
>Please comment:
>
>H. Given metric spaces (M,d) and (M',d'), f:M->M' is continuous and
> invertible.
>C. f^(-1):M'->M is continuous.
>
>Since f is continuous, for any x in M and any eps>0 in R there exists
>a delta>0 in R such that f(B(x,eps)) is a subset of B(f(x),delta),
>where "B" is an epsilon-ball.
>
>Moreover, since f has an inverse, the image of each element of f(B(x,eps))
>is unique. In particular, only x maps to f(x), so the distance of the
>image of any other point to f(x) is non-zero. Let zeta be any positive
>number less than the greatest lower bound of the distances from f(x) of
>all members of the image of f that are not elements of f(B(x,eps)).
>
>Then, B(f(x),zeta) is a subset of f(B(x,eps)), and f^(-1)(B(f(x),zeta)
>is a subset of B(x,eps). Therefore, f^(-1) is continuous.
>
>
>One thing that worries me about this attempted proof is I'm not sure
>that I can justify the existence of zeta. It seems to me that the
>image of f should "surround" f(x) in "all directions", but I'm afraid
>that this might be an unjustified assumption.
It's a good thing you didn't prove it since, for metric spaces, and
more generally, for topological spaces, it is not always true.
As an example, consider the map f from the interval [0,2*pi) to the
standard unit circle in R^2 defined by f(t)=(cos(t),sin(t)).
f is continuous, 1-1, and onto, but f^(-1) is not continuous.
If a bijective function is such that both it and its inverse are
continuous, such a function is called a homeomorphism. Such maps
preserve all topological properties so if such a map exists, the two
spaces are essentially the same from a topological viewpoint.
A necessary condition is that there exists a bijective continuous
function from one space to the other, however as the above example
shows, that's not enough -- you also have to insure that the inverse
is continuous.
In fact, although the example shows a continuous, 1-1 function from
J=[0,1) onto the standard unit circle S in R^2, there does not exist a
homeomorphism between J and S -- can you prove this?
However, for some pairs of spaces X,Y, if there exists a continuous
bijective function f:X->Y, then f^(-1) is automatically continuous and
hence is a homeomorphism.
One such theorem is this:
Let X,Y be topological spaces. If X is compact and Y is Hausdorff,
then any continuous, 1-1 function from X onto Y is a homeomorphism.
Any metric space is Hausdorff, so as long as the domain is compact
(and hence also the image), a bijective continuous function will be a
homeomorphism, and in fact, your proof idea can probably be made to
work -- try it.
Lastly, what about a bijective continuous map from R to R? Must such a
map be a homeomorphism?
quasi
.
- References:
- Continuity of inverse of continuous function
- From: Michael Stemper
- Continuity of inverse of continuous function
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