Show that if $f: A \rightarrow B$ and $g: B \rightarrow C$ are one-one,then $g \circ f: A \rightarrow C$ is also one-one.
(N/A) To show that $g \circ f$ is one-one,we assume that $g \circ f(x_1) = g \circ f(x_2)$ for any $x_1, x_2 \in A$.
By the definition of composition of functions,this implies $g(f(x_1)) = g(f(x_2))$.
Since $g: B \rightarrow C$ is given as a one-one function,$g(y_1) = g(y_2) \implies y_1 = y_2$. Therefore,$g(f(x_1)) = g(f(x_2))$ implies $f(x_1) = f(x_2)$.
Since $f: A \rightarrow B$ is also given as a one-one function,$f(x_1) = f(x_2) \implies x_1 = x_2$.
Thus,$g \circ f(x_1) = g \circ f(x_2) \implies x_1 = x_2$,which proves that $g \circ f$ is one-one.
By the definition of composition of functions,this implies $g(f(x_1)) = g(f(x_2))$.
Since $g: B \rightarrow C$ is given as a one-one function,$g(y_1) = g(y_2) \implies y_1 = y_2$. Therefore,$g(f(x_1)) = g(f(x_2))$ implies $f(x_1) = f(x_2)$.
Since $f: A \rightarrow B$ is also given as a one-one function,$f(x_1) = f(x_2) \implies x_1 = x_2$.
Thus,$g \circ f(x_1) = g \circ f(x_2) \implies x_1 = x_2$,which proves that $g \circ f$ is one-one.
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