Show that if $f: A \rightarrow B$ and $g: B \rightarrow C$ are onto,then $g \circ f: A \rightarrow C$ is also onto.
(N/A) Let $z$ be an arbitrary element in $C$.
Since $g: B \rightarrow C$ is an onto function,there exists an element $y \in B$ such that $g(y) = z$.
Since $f: A \rightarrow B$ is also an onto function,for the element $y \in B$,there exists an element $x \in A$ such that $f(x) = y$.
Now,consider the composition $(g \circ f)(x) = g(f(x))$.
Substituting $f(x) = y$,we get $g(f(x)) = g(y)$.
Since $g(y) = z$,we have $(g \circ f)(x) = z$.
Since for every $z \in C$,there exists an $x \in A$ such that $(g \circ f)(x) = z$,the function $g \circ f: A \rightarrow C$ is onto.
Since $g: B \rightarrow C$ is an onto function,there exists an element $y \in B$ such that $g(y) = z$.
Since $f: A \rightarrow B$ is also an onto function,for the element $y \in B$,there exists an element $x \in A$ such that $f(x) = y$.
Now,consider the composition $(g \circ f)(x) = g(f(x))$.
Substituting $f(x) = y$,we get $g(f(x)) = g(y)$.
Since $g(y) = z$,we have $(g \circ f)(x) = z$.
Since for every $z \in C$,there exists an $x \in A$ such that $(g \circ f)(x) = z$,the function $g \circ f: A \rightarrow C$ is onto.
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