Hello everyone. I've tried for awhile to find a counter example for the following, it says either prove or give a counter example:(adsbygoogle = window.adsbygoogle || []).push({});

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?

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

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