Proof of f^(-1)(f(A)) >= A and f(f^(-1)(B)) <= B

[johnk]

Hello,

I'm supposed to prove this (f is some function, A is some set):
\(\displaystyle \large f^{-1}(f(A)) \ge A\)
and
\(\displaystyle \large f(f^{-1}(A)) \le A\)
and tell when it is equal (eg. \(\displaystyle \large f(f^{-1}(A)) = A\) )

The second part seems easy, it's equal when f is an injective function, right?

But somehow I can't imagine a situation where it wouldn't be equal... In the first case for example, where does the inequality appear, at f(A), or at f^(-1)(A)? I'd be very grateful for an example.

 

[galactus]

A one-to-one function fits.

If a function is one-to-one, then \(\displaystyle \L\\f^{-1}(f(x))=x, \;\ f(f^{-1}(x))=x\)

Example:

\(\displaystyle \L\\f(x)=3x-5\)

Inverse:

\(\displaystyle \L\\y=3x-5\)

\(\displaystyle \L\\x=\frac{y+5}{3}\)

\(\displaystyle \L\\x=f^{-1}(y)=\frac{y+5}{3}\)

\(\displaystyle \L\\f^{-1}(x)=\frac{x+5}{3}\)

Now, we must verify:

\(\displaystyle \L\\f^{-1}(f(x))=f^{-1}(3x-5)=\frac{(3x-5)+5}{3}=x\)

\(\displaystyle \L\\f(f^{-1}(x))=f(\frac{x+5}{3})=3(\frac{x+5}{3}-5=x\)

Is that what you were asking?.

 

[johnk]

Thank you for your reply.

Yes, I understood that for an injective function (one-to-one function = injective function as I understand it) \(\displaystyle \large f^{-1}(f(X))=X\) will be true, but I wanted to know when is \(\displaystyle \large f^{-1}(f(X)) > X\) or \(\displaystyle \large f(f^{-1}(X)) < X\). The question wording seems to imply this is true in some cases (it won't be a one-to-one function obviously), otherwise why would he (our professor) not just write "=" instead of "<=" and ">=" respectively?

 

[galactus]

\(\displaystyle \L\\f(x)=x^{2}\) is a nice example of no inverse.

For if x=2, then the inverse of f at 4 should get us back to 2.

But, if x=-2, the inverse of f at 4 gets us back to 2, not -2. So, no inverse.

 

[pka]

I wonder if johnk has some notations confused.
We do not usually write \(\displaystyle f^{ - 1} \left( {f\left( A \right)} \right) \ge A\) unless we are dealing with cardinality of the sets. This makes sense for finite sets. However, in a beginning course there are really just two infinite cardinalities.

Is it possible that johnk is talking about subsets such as \(\displaystyle A \subseteq f^{ - 1} \left( {f\left( A \right)} \right)\)?

There is a standard set of theorems using this idea.
It is always true that: \(\displaystyle A \subseteq f^{ - 1} \left( {f\left( A \right)} \right)\).
If f injective then the two are equal.

It is always true that: \(\displaystyle f\left( {f^{ - 1} \left( A \right)} \right) \subseteq A\).
If f surjective then the two are equal.

 

[johnk]

I wonder if johnk has some notations confused.
We do not usually write \(\displaystyle f^{ - 1} \left( {f\left( A \right)} \right) \ge A\) unless we are dealing with cardinality of the sets. This makes sense for finite sets. However, in a beginning course there are really just two infinite cardinalities.

Is it possible that johnk is talking about subsets such as \(\displaystyle A \subseteq f^{ - 1} \left( {f\left( A \right)} \right)\)?

This does make a lot of sense, but I'm fairly certain that our prof wrote what I wrote here (less or equal and greater or equal instead of subsets).

There is a standard set of theorems using this idea.
It is always true that: \(\displaystyle A \subseteq f^{ - 1} \left( {f\left( A \right)} \right)\).
If f injective then the two are equal.

It is always true that: \(\displaystyle f\left( {f^{ - 1} \left( A \right)} \right) \subseteq A\).
If f surjective then the two are equal.

I do think this might be what was meant though, and this is how I understood the formulas, but I still don't get how is that supposed to work.

Let's take the first theorem you wrote:
\(\displaystyle A \subseteq f^{ - 1} \left( {f\left( A \right)} \right)\)

I made a simple example of how I imagine that (modified wikipedia picture):

The function is not injective, so one arrow is lost in the process... shouldn't the theorem be \(\displaystyle f^{ - 1} \left( {f\left( A \right)} \right) \subseteq A\)?

And I can't imagine the how would an example of the second theorem look like at all :-( (when it isn't equal).

Do those theorems have some name so I can look for more info about them?

 

[pka]

The function is not injective, so one arrow is lost in the process... shouldn't the theorem be \(\displaystyle f^{ - 1} \left( {f\left( A \right)} \right) \subseteq A\)? Do those theorems have some name so I can look for more info about them?

No actually both are stated correctly.
I will all the short proofs:
It is always true that: \(\displaystyle U \subseteq f^{ - 1} \left( {f\left( U \right)} \right)\).
If f injective then the two are equal.
\(\displaystyle x \in U\quad \Rightarrow \quad f(x) \in f(U)\quad \Rightarrow \quad x \in f^{ - 1} \left( {f(U)} \right)\)

It is always true that: \(\displaystyle f\left( {f^{ - 1} \left( W \right)} \right) \subseteq W\).
If f surjective then the two are equal.
\(\displaystyle \begin{array}{rcl}
y \in f\left( {f^{ - 1} (W)} \right) & \Rightarrow & \quad \underbrace {\left( {\exists t \in f^{ - 1} (W)} \right)}_{}\left[ {f(t) = y} \right] \\
& \Rightarrow & \quad y = f(t) \in W \\
\end{array}\)

Look at your own example.
\(\displaystyle f\left\{ {1,4} \right\} = \{ d,c\} \quad \quad f^{ - 1} \left( {f\left\{ {1,4} \right\}} \right) = \left\{ {1,3,4} \right\}\)