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EXERCISE 6.1 · Q46

Q.Show that if f:A→Bf:A\to B and g:B→Cg:B\to C are onto, then g∘fg\circ f is also onto.

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Let f:A→Bf:A\to B and g:B→Cg:B\to C both be onto. We must show g∘f:A→Cg\circ f:A\to C is onto.

Let z∈Cz\in C be arbitrary. Since g:B→Cg:B\to C is onto, there exists y∈By\in B with g(y)=zg(y)=z.

Since f:A→Bf:A\to B is onto, there exists x∈Ax\in A with f(x)=yf(x)=y.

Then (g∘f)(x)=g[f(x)]=g(y)=z(g\circ f)(x)=g[f(x)]=g(y)=z. …

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