Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I have $$\begin{align} X &= \{a, b, c, d, e\} \\ Y &= \{g, h, i\}\\ Z &= \{j, k, l, m, n, o \}\\ \end{align}$$ how do I write an onto function from $X$ to $Y$? Also, how do I write an one to one function from $X$ to $Z$?

Am I right to say that for the one to one function, $X$ to $Z = \{(a,j), (b,k), (c,l), (d,m), (e,n)\}$?

As for the onto function, I have to write out $4! = 24$?

Edit: onto function from $X$ to $Y = \{(a,g),(b,h), (c, i), (d,g), (e, h)\}$ something like this? But as for the one to one function. There's $5$ elements in $X$ and $6$ elements in $Z$, so it's ok leave one out?

share|improve this question

2 Answers 2

You have given one example of a one to one function $X\to Z$. There are others, but one is sufficient.

There are $540$ onto functions $X\to Y$, I'm not sure where you got $4!$ from. Again, it seems as if you need only find one example. If you're writing the functions in terms of sets, you need some set of $5$ pairs such that each element of $Y$ is the second entry in one of those pairs. The example you added satisfies this.

share|improve this answer
    
Are you sure about the $90$? –  Thomas Feb 4 '13 at 15:50
    
No - thanks! But it shouldn't be $60$ either, when you've hit the $3$ elements of $Y$ there are still $2$ elements of $X$ to be mapped to $Y$, and $3$ possible images for each of them, so there are $540$ maps. –  Matt Pressland Feb 4 '13 at 15:54
    
You are right. Together we found the right one :) –  Thomas Feb 4 '13 at 15:55
1  
I find that it usually takes two people to do combinatorics correctly! –  Matt Pressland Feb 4 '13 at 15:59

Your one-to-one function is correct. The idea is just that each element in $Z$ is only reached once and you have defined the function on each element in the domain $X$.

For the onto function that you have suggested in your edit, that is also correct. You reach all the elements and you have defined the function for each element in $X$.

As for the number of onto functions from $X$ to $Z$ note that you have to reach each element in $Y$ and you have to map each element in $X$ to something. So for each element in $Y$ you think about the number of ways that this element can be reached. For example you need to map somethin to $g$ ($5$ ways to do that) then you have will have $4$ ways to map to $h$ and after picking that, you have $3$ ways to map to $i$. Now you have then $5\cdot 4\cdot 3 = 60$ onto functions that map three elements in $X$ to something in $Y$. However, you also need to map the last two elements in $X$ to something in $Y$. Each element can be mapped to any of the three elements in $Y$. So in all you get $5\cdot 4\cdot 3 \cdot 3\cdot 3 = 540$ onto maps.

If you want to make an one-to-one function from $X$ to $Z$ you can leave out an element from $Z$. The important thing is that each element in the range is only reached once. In fact what you see is that you can't make an onto function from $X$ to $Z$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.