# [SOLVED] A few intro to advanced math topics

• May 19th 2007, 11:56 PM
welcometo1984
[SOLVED] A few intro to advanced math topics
Ok, so I have a few problems that I need to work on and have some progress made by Monday. The problem is that I need to work 30+ hours this weekend. So thanks in advance to anyone who assissts in solving, offers some advice, or views this topic

1) Let f: X--> Y and let A,B be subsets of Y. Prove that f^-1(A) - f^-1(B)= f^-1 (A-B).

Kind of a fun problem. Just show containment both ways.

2) Let f: X-->A and g: Y-->B. Define a function f x g by (f x g)(x,y)= (f(x),g(y)) for all (x,y) as elements in X x Y.

note that f x g is f cross g, not f times g.

Prove the following
a) If f and g are one-ton-one, then f x g is one-to-one.

b) If f maps onto A and g maps onto B then f x g maps on A x B

3) True or False? If false give a specific counter example (So indicate exactly what the sets and functions involved are) Assume f: X --> Y and
g: Y--> Z

a) If F^-1(Y) = X, then f maps onto Y

b) If g composed with f maps onto Z, then f maps onto Y

Again, thank you very much
• May 25th 2007, 04:20 PM
Rebesques
Quote:

f^-1(A) - f^-1(B)= f^-1 (A-B).

For $\displaystyle f^{-1}(A) - f^{-1}(B)\subseteq f^{-1}(A-B)$:

Let $\displaystyle a\in f^{-1}(A) - f^{-1}(B)$.
Then
$\displaystyle a\in f^{-1}(A)$ and $\displaystyle a\notin f^{-1}(B)$,
that is
$\displaystyle f(a)\in A$ and $\displaystyle f(a)\notin B$
so
$\displaystyle f(a)\in A-B$
or
$\displaystyle a\in f^{-1}(A-B)$.

Try the other inclusion the same way.

Quote:

If f and g are one-ton-one, then f x g is one-to-one.
Just let $\displaystyle f\times g(x,y)=f\times g(x',y')$
then
$\displaystyle (f(x),g(y))=(f(x'),g(y'))$
or
$\displaystyle f(x)=f(x')$ and $\displaystyle g(y)=g(y')$

Since now f,g are both one to one, we get.... :)

Quote:

If f maps onto A and g maps onto B then f x g maps on A x B
Let $\displaystyle (a,b)\in A\times B$. We wish to find $\displaystyle (x,y)\in X\times Y$ such that $\displaystyle f\times g(x,y)=(a,b)$.

Now since $\displaystyle a\in A$ and $\displaystyle f$ maps $\displaystyle X$ onto $\displaystyle A$, there exists some $\displaystyle x\in X$ such that $\displaystyle f(x)=a$.

Similarly, there must be some $\displaystyle y\in Y$ such that $\displaystyle g(y)=b$.

Now, check what $\displaystyle f\times g(x,y)$ is.

Ok, this is where latex officialy bores me :o

Quote:

If F^-1(Y) = X, then f maps onto Y
Let X={1} and Y={2,3}. Define f(1)=2; This f is not onto. But the inverse image of Y={2,3} through f is...

Quote:

If g composed with f maps onto Z, then f maps onto Y
With X,Y as before, define now Z={4} and g(2)=g(3)=4. Then gof (composition) maps X onto Z, but...