Since "g: B \\rightarrow C" is onto
Suppose "z \\in C" , then there exists a pre-image in B
Let the pre-image be y
Hence, "y \\in B" such that "g(y)=z"
Similarly, since "f: A \\rightarrow B" is onto
If "y \\in B" , then there exists a pre-image in A
Let the pre-image be "x"
Hence, "x \\in A" such that "f(x) =y"
Now,
"\\begin{aligned}\n\n\\text { gof : } A \\rightarrow C \\\\\n\n \\begin{aligned}\n\n\\text { gof } &=g(f(x)) \\\\\n\n&=g(y) \\\\\n\n&=z\n\n\\end{aligned}\n\n\\end{aligned}"
So, for every "x" in A, there is an image "z" in C . Thus, "gof" is onto.
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