# Composites of injections/surjections/bijections

1. May 8, 2007

### ricardianequiva

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

Consider arbitrary sets A, B, C and D with arbitrary functions:
f:A-->B, g:B-->C, h:C-->D. We define a composite function
h o g o f:A-->D.
Given that h, f, and h o g o f are bijective, and g is injective, show that g is also surjective (i.e. g is bijective).
This seems almost trivial to me, but my TA says that it requires proving.

Last edited: May 8, 2007
2. May 8, 2007

### StatusX

I don't understand the question. Are there any conditions on g besides it being injective? You can always compose functions between sets like that, so that doesn't tell you anything.

3. May 8, 2007

### ricardianequiva

No there aren't any other conditions.

4. May 8, 2007

### StatusX

Here's a similar question: Let x,y,z be integers, and define x+y+z. Show that if x and z are even, so is y. Do you see why you need more information? And what made you think it was trivial?

5. May 8, 2007

### ricardianequiva

O I made a mistake in the question, we are given that h o g o f is bijective.

6. May 8, 2007

### StatusX

Ok, then use the fact that bijections have inverses, which are also bijections, and that the composition of bijections is a bijection.

7. May 8, 2007

### ricardianequiva

got it, thanks!

8. May 8, 2007

### ricardianequiva

One more question:
In our textbook we are given a theorem that:
If f o g is bijective then g is injective and f is surjective. I can informally see this by drawing Venn Diagrams, but how would one go about doing a formal proof.

9. May 8, 2007

### StatusX

Just go back to the definitions. For example, assume g wasn't injective. Then for some x,y, we'd have g(x)=g(y), and so f(g(x))=f(g(y)), and f o g isn't injective.