We are given two functions $f: A \to B$ and $g: B \to C$, and we are tasked with proving that if the composition $g \circ f$ (i.e., $g(f(x))$) is injective, then $g$ itself must be injective.

Proof:

Given:

• $g \circ f: A \to C$ is injective.

To Prove:

• $g: B \to C$ is injective.

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Definitions:

• Injective (one-to-one): A function $h: X \to Y$ is injective if whenever $h(x_1) = h(x_2)$, it follows that $x_1 = x_2$. In other words, different inputs lead to different outputs.

Step-by-Step Proof:

1. Assume $g \circ f$ is injective:

   This means that for all $x_1, x_2 \in A$, if $g(f(x_1)) = g(f(x_2)),$ then $x_1 = x_2$.

2. To show $g$ is injective, assume for contradiction that $g$ is not injective:

   If $g$ is not injective, then there exist $b_1, b_2 \in B$ such that: $g(b_1) = g(b_2),$ but $b_1 \neq b_2$.

3. Use $f$ to define $x_1$ and $x_2$:

   Since $f$ is a function from $A \to B$, pick some $x_1, x_2 \in A$ such that: $f(x_1) = b_1 \quad \text{and} \quad f(x_2) = b_2.$ By assumption, we know that: g(f(x_1)) = g(f(x_2)) \quad \text{(since g(b_1) = g(b_2))}.

4. Contradiction to the injectivity of $g \circ f$:

   From the injectivity of $g \circ f$, we must have $x_1 = x_2$. However, this leads to a contradiction because $f(x_1) = b_1$ and $f(x_2) = b_2$, but $b_1 \neq b_2$.

5. Conclusion:

   The assumption that $g$ is not injective leads to a contradiction. Therefore, $g$ must be injective.

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Conclusion:

If $g \circ f$ is injective, then $g$ must also be injective.

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Would you like further details or have any questions?

Here are five related questions that expand on this proof:

1. Can we conclude that $f$ is injective if $g \circ f$ is injective?
2. What happens if $f$ is not injective but $g \circ f$ is?
3. How would you prove that if $f$ is surjective and $g \circ f$ is injective, then $g$ is injective?
4. Does injectivity imply surjectivity for functions in general?
5. How does the proof change if we consider surjective functions?

Tip: When dealing with compositions of functions, analyzing the properties of each individual function can often simplify the problem.