Let f: A rightarrow B and g: B rightarrow C b | vedclass

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(B) Given that $g \circ f: A \rightarrow C$ is one-one. By definition,if $g(f(x_1)) = g(f(x_2))$,then $x_1 = x_2$ for all $x_1, x_2 \in A$. Suppose $f

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$f$ is one-one and $g$ need not be one-one

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(B) Given that $g \circ f: A \rightarrow C$ is one-one. By definition,if $g(f(x_1)) = g(f(x_2))$,then $x_1 = x_2$ for all $x_1, x_2 \in A$. Suppose $f(x_1) = f(x_2)$. Then $g(f(x_1)) = g(f(x_2))$. Since $g \circ f$ is one-one,this implies $x_1 = x_2$. Thus,$f$ must be one-one. However,$g$ does not necessarily have to be one-one. For example,if $A = \{1\}$,$B = \{2, 3\}$,$C = \{4\}$,$f(1) = 2$,$g(2) = 4$,$g(3) = 4$,then $g \circ f(1) = 4$ is one-one,but $g$ is not one-one. Therefore,$f$ is one-one and $g$ need not be one-one.

Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be any two functions and $g \circ f: A \rightarrow C$ is one-one,then

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