# One-to-one relation.

• Nov 5th 2008, 06:55 AM
tygracen
One-to-one relation.
Problem:

Let f= {(1,2), (2,3), (3,4), (4,1)} and g= {(1,3), (2,1), (3,4), (4,2),(5,1).
Find f-1 and g o f.
Is g one-to-one?
Explain
• Nov 5th 2008, 12:59 PM
clic-clac
You just have to look into what your numbers are transformed by the functions.

$\displaystyle f$ is permutation ($\displaystyle f \in S_{4}$), and it is a cycle.
$\displaystyle f=(1 2 3 4)$, so $\displaystyle f^{-1}=(4 3 2 1)$, that is to say $\displaystyle f^{-1}=\{(1,4),(2,1),(3,2),(4,3)\}$.

$\displaystyle gof$ is a function from $\displaystyle \{1,2,3,4\}$ to $\displaystyle \{1,2,3,4,5\}$.
$\displaystyle gof(1)=g(f(1))=g(2)=1$. So $\displaystyle (1,1) \in gof$, with the set notation. Just find the others elements of $\displaystyle gof$.

What can you say about $\displaystyle g(2)$ and $\displaystyle g(5)$? Then conclude.
• Nov 6th 2008, 02:10 PM
whipflip15
EDIT: Don't worry. i was confused with why you used permutations.
• Nov 6th 2008, 03:30 PM
Plato
Quote:

Originally Posted by tygracen
Problem: Let f= {(1,2), (2,3), (3,4), (4,1)} and g= {(1,3), (2,1), (3,4), (4,2),(5,1).
Find f-1 and g o f. Is g one-to-one?

This problem does not have to be done using permutations.

$\displaystyle \left( {a,b} \right) \in f^{ - 1} \mbox{ if and only if }\left( {b,a} \right) \in f$
So $\displaystyle f^{ - 1} = \left\{ {(2,1),(3,2),(4,3),(1,4)} \right\}$.

$\displaystyle \left( {a,b} \right) \in g \circ f\mbox{ if and only if }\left( {\exists x} \right)\left[ {\left( {a,x} \right) \in f\,\& \,\left( {x,b} \right) \in g} \right]$
So $\displaystyle g \circ f = \left\{ {(1,1),(2,4),(4,2),(4,3)} \right\}$.

A function is one-to-one (injective) if and only if no two pairs have the same second term.
Is that true for the function $\displaystyle g$?