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Composites of injections/surjections/bijections

by ricardianequiva
Tags: composites
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ricardianequiva
#1
May8-07, 05:51 PM
P: 14
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.
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StatusX
#2
May8-07, 06:15 PM
HW Helper
P: 2,566
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.
ricardianequiva
#3
May8-07, 06:19 PM
P: 14
No there aren't any other conditions.

StatusX
#4
May8-07, 06:47 PM
HW Helper
P: 2,566
Composites of injections/surjections/bijections

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?
ricardianequiva
#5
May8-07, 06:48 PM
P: 14
O I made a mistake in the question, we are given that h o g o f is bijective.
StatusX
#6
May8-07, 06:52 PM
HW Helper
P: 2,566
Ok, then use the fact that bijections have inverses, which are also bijections, and that the composition of bijections is a bijection.
ricardianequiva
#7
May8-07, 07:00 PM
P: 14
got it, thanks!
ricardianequiva
#8
May8-07, 07:06 PM
P: 14
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.
StatusX
#9
May8-07, 07:12 PM
HW Helper
P: 2,566
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.


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