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

MATH220 April 2011

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Question 09 (a)
Let $\displaystyle f:A\to B\quad {\text{ and }}\quad g:B\to A$ be functions so that g ○f is a surjective function. Prove that g is 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
A function $f:C\to D$ is surjective if the function hits all the elements of the arrival set D, that is if for any element d in the set D there is at least one element c in the set C such that f(c)=d.

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Solution

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 are told that the function $g\circ f$ is surjective and we want to show that the function $g$ is surjective as well. For this, let us consider an element in the image of $g$, that is an element a in the set A (since $g$ sends elements from B to elements in A). To show the surjectivity we need to show that there is at least one element b in the set B which is mapped to a by the function g. Now the function $g\circ f$ sends elements of the set A to the set A and is surjective. So there must exist an element a' in the set A such that $(g\circ f)(a')=a$ But since $(g\circ f)(a')=g(f(a'))$ we can say that the element f(a'), which is an element of the set B is the element b that we are looking for since it is mapped by g to the element a as requested. This explains why g is surjective.

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