Showing if f and g are bijective then so is g o f

  • Thread starter bonfire09
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  • #1
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Homework Statement


If ##f## and ##g## are bijective functions and ##f:A→B## and ##g:B→C## then ##g \circ f## is bijective.


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The Attempt at a Solution


Showing ##g \circ f## is one to one
Suppose that ##g\circ f(a_1)=g\circ f(a_2)=> g(f(a_1))=g((f(a_2))## Since ##g## is one to one then ##f(a_1)=f(a_2)=b##. But since f is bijective there exists ##a_1## and ##a_2## in ##A## such that ##f(a_1)=b## and ##f(a_2)=b##. Since f is one to one then ##a_1=a_2##
Showing ##g \circ f## is onto
Since ##f## is onto there exists a ##a\in A## such that ##f(a)=b## where ##b\in B##. Then ##g(b)=c## for a ##c\in C## since g is onto. Thus ##g(b)=g(f(a))=c## implies that ##g \circ f## is onto.

Would this be right?
 
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Answers and Replies

  • #2
haruspex
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Showing ##g \circ f## is one to one
Suppose that ##g(b_1)=g(b_2)##
Not a good start. You need to start with supposing ##g \circ f (a_1) = g \circ f (a_2)##
 
  • #3
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I fixed it now. Is the onto part correct?
 
  • #4
haruspex
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I fixed it now. Is the onto part correct?
Not really. You should start with some arbitrary c in C then prove there exists an a in A such that etc.
 
  • #5
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Oh I was thinking that the range of f is in the domain of g so I thought I could do that. I'm curious why is the way I did it was wrong?
 
  • #6
haruspex
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Oh I was thinking that the range of f is in the domain of g so I thought I could do that. I'm curious why is the way I did it was wrong?
You started with a b in B and found from that an a in A and a c in C. But the second of those did not require g to be onto. g is defined on B, so there would have to be a c in C. So you did not show that for any c in C there is an a in A.
 
  • #7
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I fixed the onto part.Here it is. Since ##g## is onto then for every ##c\in C## there exists a unique ##b\in B## such that ##g(b)=c##. Since ##f## is onto there exists a ##a\in A## such that ##f(a)=b##. Now we see ##c=g(b)=g(f(a))=g\circ f(a)##. Thus ##g \circ f## is onto
 
  • #8
haruspex
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I fixed the onto part.Here it is. Since ##g## is onto then for every ##c\in C## there exists a unique ##b\in B## such that ##g(b)=c##. Since ##f## is onto there exists a ##a\in A## such that ##f(a)=b##. Now we see ##c=g(b)=g(f(a))=g\circ f(a)##. Thus ##g \circ f## is onto
Good. (But it is not necessary that b is unique for this part of the proof.)
 

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