To M132 students -
Thurs 9/27/07
Here are some more sample problems for Tuesday's exam. These are related
primarily to Section 18.
1. Prove part of Theorem 18.1, i.e. be able to show that one definition of
"continuous" is equivalent to another.
2. Two forms of "pasting"
>>>>Part (f) of Theorem 18.2
>>>> The "Pasting Lemma" = Theorem 18.3
3. Page 111 Ex 3
Page 111 Ex 4
Page 111 Ex 5
Page 111 Ex 8
Page 111 Ex 9
Page 111 Ex 13
4. Note: the various parts of this "problem" (actually a bunch of possible
questions) are not connected; the spaces X,Y in part (a) are not
necessarily the same spaces as you use in part (b) etc.
Def: A function f:X --> Y is an "open map" if
for each open set U in X, f(U) is open in Y.
Note: For some authors, the word "map" means "continuous function"; our
text does not assume that. So...
(a) Give an example of spaces X, Y and a function f:X-->Y that is
continuous but is not an open map.
(b) Given an example where f is an open map but is not continuous.
Def: f:X-->Y is a "closed map" if for each closed set C in X,
f(C) is closed in Y.
(c) Give an example of f:X-->Y that is a closed map but is not
contunuous.
(d) Give an example of f:X-->Y that is continuous but is not a closed
map.
(e) Show that if f:X-->Y is a bijection, continuous, and a closed map,
then f is a homeomorphism.
(f) The function f:[0, 2pi) --> Y = the unit circle in R^2 given by
f(t) = < cos(t), sin(t) > is a continuous bijection.
Show f is not a homeomorphsm.
(remark: compare part (f) to part (e). Part (f) shows that the hypothesis
"closed" cannot be removed from (e).
(g) Give an example to show that the hypotheses "continuous" cannot be
removed from part (e).
NOTE: In #5 and 6 below, we are using just the "product topology" on the
various products.
5. In Exercise 4 page 92, you showed that the coordinate projection map f:
X x Y --> X is a continuous, open map.
(a) Prove the analogous theorem for arbitrary (i.e. possibly infinite)
products.
(b) Give an example to show that, in general (even for a product of two
spaces, X x Y), it is not a closed map.
6. When you graphed functions in Calc I did you every wonder why the
graphs looked a lot like the domains? The domain was a straight
line[segment] and the graph was another "line", just maybe wiggly. There
is a theorem lurking here:
Suppose f:X-->Y is a continuous function. In the space X x Y we define a
subspace called the "graph of f" as follows:
G = {(x,y) in X x Y | y=f(x)}.
Prove: G is homeomorphic to X.
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END OF PROBLEM SET
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