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Archangel Macsika

We are given two functions f : A → B and g : B → C. Prove that if f and g are onto, then g ◦ f is onto.

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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} \text { gof : } A \rightarrow C \\ \begin{aligned} \text { gof } &=g(f(x)) \\ &=g(y) \\ &=z \end{aligned} \end{aligned}

So, for every x in A, there is an image z in C . Thus, gof is onto.

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