# 1-1 and onto functions

1. Sep 25, 2013

### Lee33

1. The problem statement, all variables and given/known data

Let f: A→B be a surjection and g:B→C be such that g∘f is an injection. Prove that both f and g are injections

2. Relevant equations

3. The attempt at a solution

Suppose a = a' then g(f(a)) = g(f(a')) since g∘f is 1-1 we have that f(a) = f(a') hence f is 1-1.

Is this correct?

2. Sep 25, 2013

### Office_Shredder

Staff Emeritus
You assumed that a=a'and proved that f(a) = f(a'), which is simply showing that f is a function, not anything to do with whether f is 1-1. Furthermore, you went from g(f(a)) = g(f(a')) to f(a) = f(a') which seems to assume that g is injective already.

You never used the fact that f is a surjection, which should also tip you off that you missed something.

3. Sep 25, 2013

### Lee33

Let me redo.

Suppose a,a' ∈ A and that f(a) = f(a') then g(f(a)) = g(f(a')) since g∘f is an injection then we have a = a' therefore f is injective.

Is this correct?

4. Sep 25, 2013

### Office_Shredder

Staff Emeritus
That looks better. Any ideas on what to do for g?

5. Sep 25, 2013

### Lee33

Thanks! For g:

Let b, b' ∈ B. Then since f is onto there exists an a, a' ∈ A such that f(a) = b and f(a') = b'. Suppose g(b)=g(b') then we get g(b) = g(f(a)) = g(f(a'))= g(b') since g∘f is 1-1 then a = a' hence f(a) = f(a') therefore b = b' which means g is 1-1.

Is this correct?

6. Sep 25, 2013

### Office_Shredder

Staff Emeritus
That looks good to me.

7. Sep 25, 2013

### Lee33

Thank you very much!