# Function counterexample

1. Jan 31, 2009

### foxjwill

1. The problem statement, all variables and given/known data
Let $$f: A\to B$$. I'm trying to find a function $$g: B\to C$$ such that $$g$$ is not $$1-1$$ but $$g\circ f$$ is.

The original assignment (which I've completed) was to prove that for all functions $$f: A\to B$$ and $$g: B\to C$$, if $$g\circ f$$ is $$1-1$$, then so is $$f$$. However, in the process of completing the assignment, I tried (out of curiosity) to find g's that weren't 1-1. But I couldn't.

2. Jan 31, 2009

### Preno

f: n -> (n,n)
g: (m,n) -> m
gf = identity, neither f nor g is one-to-one

3. Jan 31, 2009

### VeeEight

Have you thought about why it may not be possible? If the composite is defined then $$g\circ f$$ = g(f(x)) where x is in A. If f(x) is bijective then g(f(x)) = g(B)

4. Jan 31, 2009

### foxjwill

First of all, when specifying a function (and worrying about its properties), without the domain and codomain, you can't really tell much.

But assuming you have f: R -> R^2, and g: R^2 -> R, then f actually is one to one.

edit:
But since g isn't one-to-one, you have answered my question. So, thanks!

5. Jan 31, 2009

### Preno

The domain is any arbitrary set A with more than 1 element, the codomain is A x A.