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I am unable to get the point that how does it prove that it is one to one?

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______ #1: "$g$ is one to one"

______ #2: "$f$ is one to one"

The meaning of "$g$ is one to one" is that $g(x_1)=g(x_2)\implies x_1=x_2$ for any $x_1,x_2\in B$.

In your case, we have assumed that $$(g\circ f)(a_1)=(g\circ f)(a_2).$$ In other words, $$g(f(a_1))=g(f(a_2)).$$ Thus, we have two elements $f(a_1),f(a_2)\in B$ such that $g(f(a_1))=g(f(a_2))$. By the assumption that $g$ is one to one, this implies that $f(a_1)=f(a_2)$.

Then use the same reasoning to conclude that $a_1=a_2$ because $f$ is one to one.

Thus, $(g\circ f)(a_1)=(g\circ f)(a_2)\implies a_1=a_2$ for any $a_1,a_2\in A$. Therefore the function $(g\circ f):A\to C$ is one to one.

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How did you came up with,"The meaning of "g is one to one" is that g(x1)=g(x2)⟹x1=x2 for any x1,x2∈B?" The meaning of one to one to me is that there will be different outputs for all inputs. How is x1= x2 then? –  Fahad Uddin Sep 11 '11 at 12:38
    
A more precise way of saying it is "there will be different outputs for different inputs". Thus, if inputs $x_1$ and $x_2$ give the same output (i.e. $g(x_1)=g(x_2)$), then they must be the same input (i.e. $x_1=x_2$). –  Zev Chonoles Sep 11 '11 at 15:16

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