Onto and on to one functoins, if a f is onto must g be onto?

[mr_coffee]

Hello everyone. I've tried for awhile to find a counter example for the following, it says either prove or give a counter example:

If f; x->y and g: y -> z are functions and g o f is onto, must both f and g be onto? Prove or give a counter example.

for somthing to be one to one, that means all the domain in x must only point to 1 range in Y.

For somthing to be onto every Z in the range must have a co-domain. Meaning, every thing in the range must be used, you can't have any left overs. So every domain must point to a range.


My counter examples look like this:
heres my onto f:
X->Y
X = {1,2,3}
Y = {1,2,3}
the domain of 1 points to 1 in the range
the domain of 2 points to 2 in the range
the domain of 3 poitns to 3 in the range
( i could have mixed it up but I'm trying to make it simple)

heres my gne-to-one function:
Y->Z
Y = {1,2,3}
Z = {3,4,5}
the domain of 1 points to 3
the domain of 2 points to 4
the domain of 3 points to 5

But when I do this, they both look like onto functions, because I can't make g o f onto, if f is isn't onto.

can you find any counter examples or shall I attempt to prove it?

 

[0rthodontist]

No, you're on the right track finding counterexamples. Just play around some more. What could you do to the functions that you have defined to make one of them not onto?

 

[mr_coffee]

If i make 2 of the domains point to 1 of the same ranges that will make it not onto, but it seems to not work out as shown below

onto function g:
y->z
y = {1,2,3}
z = {1,2,3}
1->3
2->2
3->1

other function f:
x->y
x = {1,2,3}
y = {1,2,3}
1->1
2->2
3->2

g o f (1) = g(f(1)) = g(1) = 3
g o f (2) = g(f(2)) = g(2) = 2
g o f (3) = g(f(3)) = g(2) = 2

So one of them are onto, but the g o f is not onto

Am I allowed to make one of the ranges bigger than the others? like:
function f:
x = {1,2,3,4}
y = {1, 2, 3, 4}

Then let
1->1
2->2
3->3
4->2

but don't use domain 4 for the g function?