We are asked to prove that a function $f: A \to B$, where $|A| = |B|$ (i.e., the cardinalities of sets $A$ and $B$ are equal), is one-to-one if and only if it is onto.

Definitions:

• One-to-one (injective): A function $f$ is one-to-one if $f(a_1) = f(a_2)$ implies $a_1 = a_2$. In other words, different elements in $A$ map to different elements in $B$.
• Onto (surjective): A function $f$ is onto if for every $b \in B$, there exists some $a \in A$ such that $f(a) = b$. In other words, every element of $B$ is hit by some element from $A$.

Proof Outline:

We will prove this in two parts:

1. If $f$ is one-to-one, then $f$ is onto.
2. If $f$ is onto, then $f$ is one-to-one.

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Part 1: If $f$ is one-to-one, then $f$ is onto.

• Assume that $f$ is injective (one-to-one).
• Since $f$ is injective, no two elements in $A$ map to the same element in $B$. Therefore, $|f(A)| = |A|$, meaning the image of $A$ under $f$ has the same cardinality as $A$.
• But $|A| = |B|$ by assumption, so $|f(A)| = |B|$.
• Therefore, $f(A) = B$, meaning that $f$ is onto.

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Part 2: If $f$ is onto, then $f$ is one-to-one.

• Assume that $f$ is surjective (onto).
• Since $f$ is onto, every element of $B$ has a preimage in $A$, meaning $f(A) = B$.
• Because $|A| = |B|$, the number of preimages in $A$ must be exactly equal to the number of elements in $B$.
• If $f$ were not injective, there would be at least two distinct elements in $A$ mapping to the same element in $B$, which would contradict the fact that $|A| = |B|$.
• Therefore, $f$ must be injective, or one-to-one.

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Conclusion:

Since we have shown that if $f$ is injective, it is also surjective, and if $f$ is surjective, it is also injective, we conclude that $f$ is one-to-one if and only if it is onto.

Let me know if you'd like further details or have any questions.

Here are some related questions to deepen your understanding:

1. What happens if $|A| > |B|$? Can $f$ still be one-to-one or onto?
2. How does the proof change if $|A| \neq |B|$?
3. Can you find a concrete example of a bijection (both one-to-one and onto) between two sets?
4. What does the property of a function being both one-to-one and onto imply about its inverse?
5. How does the proof differ for infinite sets $A$ and $B$?

Tip: A bijective function (both one-to-one and onto) always has an inverse function that undoes the mapping.