To prove that if $g \circ f$ is one-to-one, then $f$ must also be one-to-one, we proceed as follows:

Given:

• $f : A \to B$ and $g : B \to C$ are functions.
• $g \circ f$ is one-to-one (injective).

Goal:

• Show that $f$ is one-to-one.

Proof:

1. Assume $f$ is not one-to-one. This would mean there exist distinct elements $a_1, a_2 \in A$ such that $a_1 \neq a_2$ but $f(a_1) = f(a_2)$.

2. Since $g \circ f$ is the composition of $f$ and $g$, we have: $(g \circ f)(a_1) = g(f(a_1)) \quad \text{and} \quad (g \circ f)(a_2) = g(f(a_2))$

3. But from step 1, we have $f(a_1) = f(a_2)$, so: $g(f(a_1)) = g(f(a_2))$ which implies that $(g \circ f)(a_1) = (g \circ f)(a_2)$.

4. Since $a_1 \neq a_2$ but $(g \circ f)(a_1) = (g \circ f)(a_2)$, this contradicts the assumption that $g \circ f$ is one-to-one.

5. Therefore, our assumption that $f$ is not one-to-one must be false.

Conclusion:

Since assuming $f$ is not one-to-one leads to a contradiction, we conclude that $f$ must be one-to-one.

This completes the proof.

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. How would the proof change if $g$ were assumed to be one-to-one instead of $g \circ f$?
2. If $f$ and $g$ are both one-to-one, is $g \circ f$ necessarily one-to-one as well?
3. What happens if $g \circ f$ is onto (surjective) instead of one-to-one? Does this imply anything about $f$ or $g$?
4. Can you provide an example of functions $f$ and $g$ where $g \circ f$ is one-to-one, and $f$ is also one-to-one?
5. How does this proof change if we consider $g \circ f$ to be a many-to-one function?

Tip: In proofs involving functions, carefully consider definitions like injective (one-to-one) and surjective (onto), and focus on the implications of these properties on the function composition.