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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.

## Homework Equations

## 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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