# Sets, Relations and Functions

1. Jan 31, 2010

### L²Cc

1. The problem statement, all variables and given/known data
List all the functions from {1,2,3} to {1,2} representing each function as an arrow
diagram. Which of these functions are (a) injective, (b) surjective, (c) bijective? For
each surjective function write down a right inverse.

2. Relevant equations

3. The attempt at a solution
The only one I can think of is,
f(x) = x - 1, where X is all real numbers,
and in the directions it specifically asks for functionS...

2. Jan 31, 2010

### Staff: Mentor

You're think in terms of formulas of functions, when that isn't the direction you should go. In fact, function formulas have no role to play in this problem. Recall that a function is simply a mapping of members in one set to those of another set such that no member of the first set gets mapped to multiple members of the second set.

Look at your problem description again. You are supposed to be drawing arrows from the members of the first set, {1, 2, 3}, to the second set, {1, 2}. Look in your textbook or notes for the arrow diagrams that your problem refers to.

3. Jan 31, 2010

### L²Cc

oh right right!
in other words,
f(1) = 1
f(2) = 2
f(3) = 2
functions f(2) and f(3) would have been injections had a and b in the following f(a) = f(b)were equal which is not the case here? Surjections have have the property that for every y in the codomain there is an x in the domain such that ƒ(x) = y, f(1) and f(2) surjections? Not f(3) because it maps to the same element in the range as f(2) ?

Thank you.

4. Jan 31, 2010

### VeeEight

Your f is surjective since every element in the codomain are in the image of f. It is not injective since it f(2)=f(3), and therefore is not a bijection.

Recall that a function is injective or one to one if all elements in the domain map to unique elements of the codomain. A function is surjective or onto if all elements from the codomain are covered. An a bijection is a function that is both an injection and a surjection.

Last edited: Jan 31, 2010
5. Jan 31, 2010

### L²Cc

Makes sense. Thank you!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?