# If f: A →</

We have, f: A B and g: B C are onto functions.

Now, we need to prove: gof: A C is onto.

let y C, then

gof (x) = y

g(f(x)) = y ……(i)

Since g is onto, for each element in C, there exists a preimage in B.

g(x)=y ……(ii)

From (i) & (ii)

f(x)=x

Since f is onto, for each element in B there exists a preim age in el

f(x)=x ……(iii)

From (ii)and(iii) we can conclude that for each y C, there exists a preimage in A such that gof(x) = y

gof is onto.

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