Let A, B and C be sets and let f:A rightarrow B and g: B rightarrow C be functio

Question

Let A, B and C be sets and let f:A rightarrow B and g: B rightarrow C be functions. Prove that, if each of f and g is one-to-one, them g of is one-to-one. Prove that. If each of f and g is onto, then g o f in onto.

Explanation / Answer

(a)if f is 1-1 then we may write f(x)=f(y) => x=y (1) similarly g(x)=g(y) => x=y (2) now consider gof(x)=gof(y) g(f(x))=g(f(y)) g(x)=g(y) by (!) x=y by (2) hence gof is 1-1 (b) It's a matter of putting everything in the right order. We want to show that (g o f) is onto. Hence, given c in C, we need to find a in A such that (g o f)(a) = g(f(a)) = c. ------------ To do this, what can we do with our element c? Only the individual map g deals with set C. *Since g is onto, there exists b in B such that g(b) = c. Now, we have an element in B; can we now get back to A? Yes we can; use f. *Since f is onto, there exists a in A such that f(a) = b. (This is the 'a' we want!) ----------------- Here's the proof (without annotation): Given c in C: Since g is onto, there exists b in B such that g(b) = c. Next, since f is onto, there exists a in A such that f(a) = b. So, (g o f)(a) = g(f(a)) = g(b) = g(c). Therefore, (g o f): A ? C is onto. -------------- I hope this helps!

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