Functions and Inverses

**tukeywilliams** · July 9th 2007, 12:30 PM

I am not exactly sure what these functions mean:

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

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

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

So if $f$ is a bijection with inverse $f^{-1}$ , then $f(x) = y_0$ if and only if $x = f^{-1}(y_0)$ so that $\overleftarrow{f}(\{y_0 \}) = \{f^{-1}(y_0) \}$ . So this implies that $\overleftarrow{f}(\{y_0 \}) \neq \{f^{-1}(y_0) \}$ sometimes?

**Plato** · July 9th 2007, 12:36 PM

Originally Posted by tukeywilliams

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

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

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

No, they are really two new functions related to the function $f$ .
The two new functions are mapping of subsets to subsets.
Whereas $f$ maps set X to set Y.

**tukeywilliams** · July 9th 2007, 12:52 PM

Yes, but why do we need these functions? Are they only useful for figuring out whether functions have inverses?

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

**ThePerfectHacker** · July 9th 2007, 12:58 PM

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 $\phi: G \mapsto G'$ then $\phi ^{-1} [\{e\}]$ where $\{e'\}\subseteq G'$ is called the kernel it is extremely important concept.

Have you even taken linear algebra? Given a linear transformation $T: V\to V'$ then $\ker T$ is defined exactly like above $T^{-1} [ \bold{0}]$ where $\bold{0}$ is the zero vector of $V'$ . This is another important concept.

**tukeywilliams** · July 9th 2007, 12:59 PM

Yes I have taken linear algebra.

**ThePerfectHacker** · July 9th 2007, 01:01 PM

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.

**tukeywilliams** · July 9th 2007, 01:07 PM

Yeah the kernel of a linear transformation is the set of vectors v such that $T(v) =0$ , 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 $\{0 \}$ and vice versa.

**Plato** · July 9th 2007, 01:23 PM

Originally Posted by tukeywilliams

Yes, but why do we need these functions?

In other words, if $\overleftarrow{f}(\{y_0\}) \neq \{f^{-1}(y_0)\}$ then $f$ 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 $f$ is a function from A to B then $f$ is also a relation from A to B. Now it is true then that $f^{-1}$ is also a relation from A to B BUT it may not be a function.

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

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