Re: Topological graphs

On Mon, 01 Jun 2009 03:33:26 EDT, Jose Capco <jcapco@xxxxxxxxx> wrote:

Then h(x) = (p1.h(x), p2.h(x)) = (p1.h(x),
f(p1.h(x)))
. . where p1 and p2 are the 1st and 2nd
projections.

As p2.h is continuous, so is f.p1.h.
As p1.h is a quotient map, f is continuous.
?QED.

There is exactly one step in it which is
not clearly correct: can you see which?

No. Claiming p1.h is a quotient?

I think that is a candidate for suspicion. You could
also trace
the single steps of your proof on your
counterexample.

--
Marc

Well... let's see

p1 is a projection map, so its an open map.
h is a homeomorphism, so it is an open map
p1.h is a quotient map so it is surjective and open.

Let U be an open set in Y. Define V = f^{-1}(U).
and define (p1.h)^{-1}(V) = V_0 .. V_0 is open because f.p1.h is continuous. Now p1.h is an open, surjective continuous map. So we get p1.h(V_0) =V is open.

So that part was ok.

I'm beginning to like the proof. What I don't understand quite is the "counterexample" he provided. In his counterexample, he had an h: X -> G_f .. but
p2.h(X) = {0} .. yet p2(G_f) = {0,1,3,5, ...}

So William, take a look at your "counterexample" again.

The counterexample is obviously correct - look at a picture of the
graph, it's obviously homeomorphic to X. There was a typo (or
something) in the formulas William gave for the definition of
the homeomorphism; the formulas he wrote down don't map
X to G_f.

David C. Ullrich

"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
.

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