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

MATH220 April 2011

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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.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

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