Question

Let \(f: X \rightarrow Y, g: Y \rightarrow Z\) and \(h: Z \rightarrow U\). Prove that for all \(T \in P(U), \bar( h \circ g \circ f)^{-1}(T)=\bar f^{-1}(\bar g^{-1} (\bar h^{-1}(T)))\).

Answer 1

-.-

Let \(f: X \rightarrow Y, g: Y \rightarrow Z\) and \(h: Z \rightarrow U\). Prove that for all \(T \in P(U), \bar( h \circ g \circ f)^{-1}(T)=\bar f^{-1}(\bar g^{-1} (\bar h^{-1}(T)))\).

Answer 2

let's see.. why can't I just use the distributive law to solve my iissues? lol that' snot the case

Answer 3

what is \( P(U)\)?

Answer 4

so f has an element x with element y g has an element y with element z and h has an element z with element u...

Answer 5

that's supposed to be POWERSET(U)

Answer 6

so this is preimage not inverse?

Answer 7

yes ^^

Answer 8

what if we use some lemma.. which states that we let X and Y be nonempty sets and let f: x -> Y be a function...

If f is an injection, then there exists a surjection g : y -x such that g o f =Ix
If f is a surjection, then there exists an injection g: y ->x such that f o g = Iy ?????

Answer 9

How about this, take \(f\) of both sides. Then take \(g\) of both sides, then take \(h\) of both sides.

Answer 10

so that would be just ... wait do I multiply both sides of f or ?!

Answer 11

\[
f((h\circ g\circ f)^{-1}(T)) = g^{-1}(h^{-1}(T))
\]Then\[
(g\circ f)((h\circ g\circ f)^{-1}(T)) = h^{-1}(T)
\]

Answer 12

Basically: \[
\color{red}{(h\circ g\circ f)^{-1}(T)}= \color{blue}{f^{-1}(g^{-1}(h^{-1}(T)))}
\]Then\[
f[\color{red}{(h\circ g\circ f)^{-1}(T)}] = f[\color{blue}{f^{-1}(g^{-1}(h^{-1}(T)))}]
\]The RHS simplifies to: \[
f[(h\circ g\circ f)^{-1}(T)] = g^{-1}(h^{-1}(T))
\]

Answer 13

...

Answer 14

wait how did you get it so fast? like when did the ff^-1 become just g^-1h^-1 T?

Answer 15

\[
f(f^{-1}(\ldots )) = \ldots
\]

Answer 16

In this case \(\ldots = g^{-1}(h^{-1}(T))\)

Answer 17

so does the ff^-1 cancel out?

Answer 18

?! what properties of functions did you use?!

Answer 19

This went too quickly... so what did you do for take f on both sides?

Answer 20

Start out with \(T=T\)

Answer 21

:S T = T is this from a property or a lemma or letting T equal to \[f((h\circ g\circ f)^{-1}(T)) = g^{-1}(h^{-1}(T)) \]

Answer 22

No, \(T=T\) is something that we know is true.

Answer 23

Thus we can start with it.

Answer 24

Actually, maybe we should start with: \[
h(g(f(T))) = (h\circ g\circ f)(T)
\]Does this at least make sense?

Answer 25

Divide by 0.

Answer 26

Now, we put \(h^{-1}(\ldots)\) on both sides: \[
h^{-1}[h(g(f(T)))] = h^{-1}((h\circ g\circ f)(T))
\]Cant we use \(h^{-1}(h(\ldots)) = \ldots\)

Answer 27

@primeralph divide by 0?! really I thought that's undefined already?!

Answer 28

@UsukiDoll Do you understand what I'm doing?

Answer 29

we're starting from h first... then take the h^-1 on both sides

Answer 30

We get \[
g(f(T)) = h^{-1}((h\circ g\circ f)(T))
\]

Answer 31

h^-1 (h) just disappears... from the left side... :/

Answer 32

Eventually you get \[
T = f^{-1}(g^{-1}(h^{-1}((h\circ g\circ f)(T))))
\]

Answer 33

Let \(X = (h\circ g\circ f)(T)\) then \(T = (h\circ g\circ f)^{-1}(X)\)

Answer 34

Substitute and get: \[
(h\circ g\circ f)^{-1}(X) = f^{-1}(g^{-1}(h^{-1}(X)))
\]

Answer 35

so first we're taking h^-1 both sides... then maybe g^-1 both sides and f^-1 both sides to finally get to where we are

Answer 36

Yes.

Answer 37

And... I probably should not have used \(X\) because it is already defined here... so instead...

Answer 38

use ... B?

Answer 39

Change \(X\) to be \(T\). Change \(T\) to be some \(A\in \mathcal P(X)\)

Answer 40

That way instead of ending with \(X\) you end with \(T\)s, which is less confusing.

Answer 41

Do you understand it?

Answer 42

if I do it backwards starting with h^-1 on both sides then I do...but for forward... should I take f^-1 on both sides first followed by g^-1 and then h^-1

Answer 43

You technically can't do it forwards, because you would be assuming the thing you're trying to prove is true.

Answer 44

However, sometimes doing this forwards gives us an idea.

Answer 45

Sometimes, I feel you won't be satisfied unless I write a verbatim proof.

Answer 46

oh I see I'll start on the left hand side of the equation on this one... doing h^-1 then g^-1 and f^-1

Answer 47

on both sides of course ^^

Answer 48

either way since professor I love to grade harsh is the teach....no matter what I do I still have some points knocked out

Answer 49

To be more rigorous, you'd start with \[
\forall A\in \mathcal P(X) \quad h(g(f(A))) = (h\circ g\circ f)(A)
\]

Answer 50

alright...hmm I wonder if any definitions were used...theorem 5.5.11 Let X and Y be sets. Let f: X->Y be a function, and let g be an inverse of f. Then the following hold...
both f and g are bijections, f is an inverse of g, g is the unique inverse of f.
... uniqueness of the inverse..........

Answer 51

First we show:\[
\begin{array}{rccc}
\forall T\in \mathcal P(U) \quad& (h\circ g\circ f)(A) &=& T \\
& (h\circ g\circ f)^{-1}((h\circ g\circ f)(A)) &=& (h\circ g\circ f)^{-1}(T)\\
& A &=& (h\circ g\circ f)^{-1}(T)
\end{array}
\]Next we show: \[\begin{array}{rccc}
\forall T\in \mathcal P(U) \quad& (h\circ g\circ f)(A) &=& T \\
& h(g(f(A))) &=& T \\
& h^{-1}(h(g(f(A)))) &=& h^{-1}(T) \\
& g(f(A)) &=& h^{-1}(T)\\
& g^{-1}(g(f(A))) &=& g^{-1}(h^{-1}(T))\\
& f(A) &=& g^{-1}(h^{-1}(T))\\
& f^{-1}(f(A)) &=& f^{-1}(g^{-1}(h^{-1}(T)))\\
& A &=& f^{-1}(g^{-1}(h^{-1}(T)))\\
\end{array}
\]Then with we combine:\[
(h\circ g\circ f)^{-1}(T) = A = f^{-1}(g^{-1}(h^{-1}(T)))
\]

Answer 52

Please tell me you get it...

Answer 53

back...my pc crashed right as I was about to type something

Answer 54

ok