1. function compositions

Here's the problem: Find sets $\displaystyle A, B$ and $\displaystyle C$ and functions $\displaystyle f:A\longrightarrow B, g:B\longrightarrow C$ and $\displaystyle h:B\longrightarrow C$ for which $\displaystyle f\circ h=f\circ g$ but $\displaystyle g\not=h$.
It doesn't even have to be real example functions, just a picture would be good. I just can't think of anything. Thank you!

2. Originally Posted by sfitz
Here's the problem: Find sets $\displaystyle A, B$ and $\displaystyle C$ and functions $\displaystyle f:A\longrightarrow B, g:B\longrightarrow C$ and $\displaystyle h:B\longrightarrow C$ for which $\displaystyle f\circ h=f\circ g$ but $\displaystyle g\not=h$.
It doesn't even have to be real example functions, just a picture would be good. I just can't think of anything. Thank you!
how about $\displaystyle A = B = C = \mathbb{R}$. with $\displaystyle f(x) = x^2,~g(x) = x, \mbox{ and }h(x) = -x$

can you find another example?

3. Oh my god, I am so sorry, I typed the problem wrong. I meant $\displaystyle g\circ f$ and $\displaystyle h\circ f$! Can you still help? The part I typed was part a of the problem, and I'm stuck on part b. I'm really sorry.

4. Originally Posted by sfitz
Oh my god, I am so sorry, I typed the problem wrong. I meant $\displaystyle g\circ f$ and $\displaystyle h\circ f$! Can you still help? The part I typed was part a of the problem, and I'm stuck on part b. I'm really sorry.
how about $\displaystyle g(x) = x,~h(x) = f(x) = 1$

try to find another

5. ok, I understand, f must not be onto for it to work, so that there are elements in B that are outside the image of f. thank you!

6. Originally Posted by sfitz
ok, I understand, f must not be onto for it to work, so that there are elements in B that are outside the image of f. thank you!
can you prove that?

just kidding... well, you can if you want to

7. I did prove it! That was the other part of the question, but I couldn't prove it without being able to think of an example

8. Originally Posted by sfitz
I did prove it! That was the other part of the question, but I couldn't prove it without being able to think of an example
oh, ok. that's fine then