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Question 04 (b) 

Let : and be functions so that is an injection. Prove that need not be injective. 
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 

From the previous part of this problem, we know that is an injection. This means that . If indeed it were the case that then with a bit of effort you could convince yourself that should be an injection. If a counter example is to exist, it must exist with . Try to find one using small finite sets. 
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. Let , , be the function sending 1 to 2 and let be the function sending all of to 1. Then is the function sending 1 to itself and hence is injective. However is not injective since it sends both 2 and 3 to the same element. This completes the counterexample. 