- #1

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- Homework Statement
- Suppose ##A## and ##B## are finite sets and ##f: A \longrightarrow B ##. Prove that if ##|A| = |B| ## then ##f## is one to one if and only if ##f## is onto.

- Relevant Equations
- Definition of one to one and onto function

Since this is bi-conditional, we have two directions to prove. Given is ##|A| = |B|= n ##. Now suppose that ##f## is one to one. This means that given ##a_1, a_2 \in A## , if we have ##a_1 \ne a_2## then we have ##f(a_1) \ne (a_2) ##. So, we can not have any two elements in ##A## mapped to a single element in ##B##. This means that range of ##f## must have ##n## elements. But we are given that ## |B| = n##. So range of ##f## must be same as ##B##. This means that all elements of ##B## have some pre-image in ##A##. Which proves that ##f## is an onto function.

This was forward direction. I want to know if the reasoning is right here.

Thanks

This was forward direction. I want to know if the reasoning is right here.

Thanks