# Discrete Math: Onto

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## 2 Replies - 226 Views - Last Post: 05 October 2014 - 10:34 PM

### #1 streek405

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# Discrete Math: Onto

Posted 05 October 2014 - 05:52 PM

Is a function onto if there are 4 domains {a, b, c, d} with 3 co-domains {x, y, z} where:
a -> x
b -> y
c -> z
d (does nothing)
?

My notes state that every element in in domain must point to an element in the co-domain, but my classmate told me that this is still onto or a surjection since d maps onto, or points to, the empty/null set.
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## Replies To: Discrete Math: Onto

### #2 macosxnerd101

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## Re: Discrete Math: Onto

Posted 05 October 2014 - 07:17 PM

Your terminology needs a bit of work. Let X = {a, b, c, d}. Then the set X is the domain. You don't have multiple domains. The elements a, b, c, and d are all in the domain. Similarly, Y = {x, y, z} is the codomain (singular).

A function f: X -> Y is defined as such. For every g in X, there exists a k in Y such that f(x) = k. So f(d) must be defined. That is, d cannot map to nothing.

An onto function is a function such that every element in Y is "hit" on the mapping. So the following is a surjection:
f(a) = f(b) = x
f(d) = z

If f(g) = x for every g in X, then the function is not onto.

I actually have a tutorial on functions you may find helpful.

### #3 streek405

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## Re: Discrete Math: Onto

Posted 05 October 2014 - 10:34 PM

macosxnerd101, on 05 October 2014 - 07:17 PM, said:

Your terminology needs a bit of work. Let X = {a, b, c, d}. Then the set X is the domain. You don't have multiple domains. The elements a, b, c, and d are all in the domain. Similarly, Y = {x, y, z} is the codomain (singular).

A function f: X -> Y is defined as such. For every g in X, there exists a k in Y such that f(x) = k. So f(d) must be defined. That is, d cannot map to nothing.

An onto function is a function such that every element in Y is "hit" on the mapping. So the following is a surjection:
f(a) = f( = x