Science:Math Exam Resources/Courses/MATH220/April 2011/Question 09 (b)

MATH220 April 2011

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[hide] Question 09 (b)
Let $\displaystyle f:A\to B\quad {\text{ and }}\quad g:B\to A$ be functions so that g ○f is a surjective function. Give an example of sets A, B and functions f, g as above such that f is not surjective.

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[show] Hint
Start with a choice of sets A and B and functions f and g such that $g\circ f$ is surjective. If f isn't surjective, you won, that's your example! If not, can you add some elements to the set B to tweak your example and obtain what you're looking for?

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[show] Solution 1
We showed in part (a) that the function g will always be surjective, this question asks us to give an example to illustrate why the function f doesn't have to be necessarily. For example, consider the case where the set A is all the natural numbers including zero: A={ 0, 1, 2, ... } and the set B is all the integers: B = { ..., -2, -1, 0, 1, 2, ... }. Now the function f just maps a natural number to itself, so $\displaystyle f(n)=n\quad {\text{for all }}n=0,1,2,\dots$ and the function g maps an integer to its absolute value, so $\displaystyle g(k)=\|k\|\quad {\text{for all }}k\in \mathbb {Z}$ Then the function $g\circ f$ is clearly surjective since it takes a natural number back to itself. That is, $(g\circ f)(a)=a$ for all $a\in A$ . But the function f isn't surjective since it doesn't map to any negative integer (no negative number has a pre-image under the function f). Notice that as shown in part (a) the function g is, and has to be, surjective.

[show] Solution 2
Another, minimal example is the following: Let A = {a}, and B = {1,2}. Then define $f:A\rightarrow B,\quad f(a)=1,\qquad g:B\rightarrow A,\quad g(1)=g(2)=a.$ Then, clearly $(g\circ f):A\rightarrow A,\quad (g\circ f)(a)=g(f(a))=g(1)=a,$ and thus $g\circ f$ is surjective. However, f is not surjective, since $2\in B$ does not have a pre-image in A under f. Notice again, that g is, and needs to be, surjective for $g\circ f$ to be surjective.