Functions and Inverses

• Jul 9th 2007, 12:30 PM
tukeywilliams
Functions and Inverses
I am not exactly sure what these functions mean:

$\displaystyle \overrightarrow{f}: \mathcal{P}(X) \rightarrow \mathcal{P}(Y)$ defined by $\displaystyle \overrightarrow{f}(A) = \{ f(x) | x \in A \}$ for $\displaystyle A \in \mathcal{P}(X)$ and

$\displaystyle \overleftarrow{f}: \mathcal{P}(Y) \rightarrow \mathcal{P}(X)$ defined by $\displaystyle \overleftarrow{f}(B) = \{x \in X | f(x) \in B \}$ for $\displaystyle B \in \mathcal{P}(Y)$.

These two functions really are $\displaystyle f$ and $\displaystyle f^{-1}$ right?

So if $\displaystyle f$ is a bijection with inverse $\displaystyle f^{-1}$, then $\displaystyle f(x) = y_0$ if and only if $\displaystyle x = f^{-1}(y_0)$ so that $\displaystyle \overleftarrow{f}(\{y_0 \}) = \{f^{-1}(y_0) \}$. So this implies that $\displaystyle \overleftarrow{f}(\{y_0 \}) \neq \{f^{-1}(y_0) \}$ sometimes?
• Jul 9th 2007, 12:36 PM
Plato
Quote:

Originally Posted by tukeywilliams
$\displaystyle \overrightarrow{f}: \mathcal{P}(X) \rightarrow \mathcal{P}(Y)$ defined by $\displaystyle \overrightarrow{f}(A) = \{ f(x) | x \in A \}$ for $\displaystyle A \in \mathcal{P}(X)$ and

$\displaystyle \overleftarrow{f}: \mathcal{P}(Y) \rightarrow \mathcal{P}(X)$ defined by $\displaystyle \overleftarrow{f}(B) = \{x \in X | f(x) \in B \}$.

These two functions really are $\displaystyle f$ and $\displaystyle f^{-1}$ right?

No, they are really two new functions related to the function $\displaystyle f$.
The two new functions are mapping of subsets to subsets.
Whereas $\displaystyle f$ maps set X to set Y.
• Jul 9th 2007, 12:52 PM
tukeywilliams
Yes, but why do we need these functions? Are they only useful for figuring out whether functions have inverses?

In other words, if $\displaystyle \overleftarrow{f}(\{y_0\}) \neq \{f^{-1}(y_0)\}$ then $\displaystyle f$ has no inverse?
• Jul 9th 2007, 12:58 PM
ThePerfectHacker
Quote:

Originally Posted by tukeywilliams
Yes, but why do we need these functions?

This inverse image function is really useful.

I do not think you taken group theory, given the homomorphism $\displaystyle \phi: G \mapsto G'$ then $\displaystyle \phi ^{-1} [\{e\}]$ where $\displaystyle \{e'\}\subseteq G'$ is called the kernel it is extremely important concept.

Have you even taken linear algebra? Given a linear transformation $\displaystyle T: V\to V'$ then $\displaystyle \ker T$ is defined exactly like above $\displaystyle T^{-1} [ \bold{0}]$ where $\displaystyle \bold{0}$ is the zero vector of $\displaystyle V'$. This is another important concept.
• Jul 9th 2007, 12:59 PM
tukeywilliams
Yes I have taken linear algebra.
• Jul 9th 2007, 01:01 PM
ThePerfectHacker
Quote:

Originally Posted by tukeywilliams
Yes I have taken linear algebra.

Do you understand what I mean by the kernel of a linear transformation. Carefully try to understand that the kernel is defined as an inverse image.
• Jul 9th 2007, 01:07 PM
tukeywilliams
Yeah the kernel of a linear transformation is the set of vectors v such that $\displaystyle T(v) =0$, i.e. the set of vectors that maps to the 0 vector.

So the inverse image function takes the kernel and maps it to the singleton set $\displaystyle \{0 \}$ and vice versa.
• Jul 9th 2007, 01:23 PM
Plato
Quote:

Originally Posted by tukeywilliams
Yes, but why do we need these functions?

In other words, if $\displaystyle \overleftarrow{f}(\{y_0\}) \neq \{f^{-1}(y_0)\}$ then $\displaystyle f$ has no inverse?

The answer to the first question is: it is a foundational issue. That is, in a course on the foundation of mathematics, logic & sets, a function is defined as a relation with additional properties. If $\displaystyle f$ is a function from A to B then $\displaystyle f$ is also a relation from A to B. Now it is true then that $\displaystyle f^{-1}$ is also a relation from A to B BUT it may not be a function.

However $\displaystyle \overleftarrow f$ is always a function from $\displaystyle P(B)$ to $\displaystyle P(A)$.