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

• mr_coffee
In summary, the question is whether both f and g must be onto if g o f is onto. The answer is no, as shown by the counterexamples given. It is possible to make one of the functions onto and the other not onto, which would result in g o f not being onto. This can be achieved by manipulating the domains and ranges of the functions.
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?

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?

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?

## What is the definition of an onto function?

An onto function, also known as a surjective function, is a function where every element in the range is mapped to by at least one element in the domain. This means that there are no "leftover" elements in the range that are not mapped to.

## What is the definition of an on to one function?

An on to one function, also known as an injective function, is a function where every element in the range corresponds to exactly one element in the domain. This means that each element in the range has a unique preimage in the domain.

## What is the relationship between onto and on to one functions?

The relationship between onto and on to one functions is that an onto function can also be an on to one function, but an on to one function does not necessarily have to be an onto function. In other words, all onto functions are on to one functions, but not all on to one functions are onto functions.

## If a function f is onto, does that mean g is also onto?

No, the fact that f is onto does not necessarily mean that g is also onto. This is because the two functions may have different domains and ranges, and therefore may have different mappings and elements in their respective ranges.

## If f is an onto function, is it possible for g to be an on to one function?

Yes, it is possible for g to be an on to one function even if f is an onto function. This is because the two functions may have different domains and ranges, and therefore may have different mappings and elements in their respective ranges. Additionally, there may be multiple elements in the domain of f that map to the same element in the range of g, making g an on to one function.

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