Injective and surjective functions But g◦ f must be bijective.

[Lineara]

Hey,
I'm looking for 2 functions f and g. One must be injective and the one must be surjective. But g◦ f must be bijective.

 

[Deleted member 4993]

Hey,
I'm looking for 2 functions f and g. One must be injective and the one must be surjective. But g◦ f must be bijective.

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

 

[pka]

I'm looking for 2 functions f and g. One must be injective and the one must be surjective. But g◦ f must be bijective.

Your actual question is not at all clear. But we do know these are true.
1 If each of f and g is injective then \(\displaystyle g\circ f\) is injective.
2 If both are surjective then \(\displaystyle g\circ f\) is surjective.
3 If \(\displaystyle g\circ f\) is injective then f is injective.
4 If \(\displaystyle g\circ f\) is surjective then g is surjective.
5 If \(\displaystyle g\circ f\) is bijective then f is injective and g is surjective.

Be careful how you use these.
It is not true that If f is injective and g is surjective then \(\displaystyle g\circ f\) is bijective.

It is true that If f is not injective or g is not surjective then \(\displaystyle g\circ f\) is not bijective.

 

[Lineara]

Hey pka!

The original statement was
If g◦ f is bijective then f and g are bijective.

Which is false (also according to #5).
I have to give an example to show that this statement wrong..

 

[Steven G]

Hey pka!

The original statement was
If g◦ f is bijective then f and g are bijective.

Which is false (also according to #5).
I have to give an example to show that this statement wrong..

Why would you not give the original statement? I'd use use the definitions to figure out f and g.
We know that if gof is bijective then f is injective and g is surjective. So find an injective only function and a surjective only function and see what happens.

 

[pka]

Your actual question is not at all clear. But we do know these are true.
5 If \(\displaystyle g\circ f\) is bijective then f is injective and g is surjective.
Be careful how you use these.
It is not true that If f is injective and g is surjective then \(\displaystyle g\circ f\) is bijective.
It is true that If f is not injective or g is not surjective then \(\displaystyle g\circ f\) is not bijective.

The original statement was
If g◦ f is bijective then f and g are bijective.

Which is false (also according to #5).

That is wrong.

If a function is bijective then it is surjective & it is injective.

 

[HallsofIvy]

Hey,
I'm looking for 2 functions f and g. One must be injective and the one must be surjective. But g◦ f must be bijective.

If that was actually the full statement of the problem, then the simplest thing to do is to take f and g to be bijective functions themselves, say, f(x)= x and g(x)= 2x.

Or was f to be injective but not surjective and g to be surjective but not injective?