This statement is true. Here is the proof:

Assume that f: A → B and g: B → C are injective functions. We want to show that the composition function g o f: A → C is also injective.

By definition, a function is injective (or one-to-one) if different inputs always produce different outputs. In other words, if x and y are elements of A and if g(f(x)) = g(f(y)), then x must equal y.

So, let's take two arbitrary elements x and y in A and assume that g(f(x)) = g(f(y)). Since g is injective, this implies that f(x) = f(y). Then, since f is also injective, this implies that x = y.

Therefore, if g(f(x)) = g(f(y)), then x = y, which means that the composition function g o f: A → C is injective. This completes the proof.

Question

This statement is true. Here is the proof:

Assume that f: A → B and g: B → C are injective functions. We want to show that the composition function g o f: A → C is also injective.

By definition, a function is injective (or one-to-one) if different inputs always produce different outputs. In other words, if x and y are elements of A and if g(f(x)) = g(f(y)), then x must equal y.

So, let's take two arbitrary elements x and y in A and assume that g(f(x)) = g(f(y)). Since g is injective, this implies that f(x) = f(y). Then, since f is also injective, this implies that x = y.

Therefore, if g(f(x)) = g(f(y)), then x = y, which means that the composition function g o f: A → C is injective. This completes the proof.

Knowee AI · Accepted Answer

This statement is true. Here is the proof:

Assume that f: A → B and g: B → C are injective functions. We want to show that the composition function g o f: A → C is also injective.

By definition, a function is injective (or one-to-one) if different inputs always produce different outputs. In other words, if x and y are elements of A and if g(f(x)) = g(f(y)), then x must equal y.

So, let's take two arbitrary elements x and y in A and assume that g(f(x)) = g(f(y)). Since g is injective, this implies that f(x) = f(y). Then, since f is also injective, this implies that x = y.

Therefore, if g(f(x)) = g(f(y)), then x = y, which means that the composition function g o f: A → C is injective. This completes the proof.

If f A —> B and g : B —> C are injectivefunction, then gof: A —> C is an injectivefunction. Prove or disprove.

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