
#1
May807, 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. 



#2
May807, 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.




#3
May807, 06:19 PM

P: 14

No there aren't any other conditions.




#4
May807, 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?




#5
May807, 06:48 PM

P: 14

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




#6
May807, 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.




#7
May807, 07:00 PM

P: 14

got it, thanks!




#8
May807, 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. 



#9
May807, 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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