Show that if `f : A ->B`and `g : B ->C`are onto, then `gof : A ->C`is also onto.

Explanation

Since h:B→C is onto

Suppose z∈C,
then there exists a pre-image in B

Let the pre-image be y

Hence, y∈B such that g(y)=z

Similarly since f:A→B is onto

If y∈B then exists a preimage in A

Let the pre image be x

Hence, x∈A such that f(x)=y

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