Question

Let $S$, $T$, $U$ be three non-void sets, where $f: S \to T$, $g: T \to U$, and the composed mapping $g \circ f: S \to U$ is defined. If $g \circ f$ is an injective mapping, then

• A. f, g both are injective.
• B. neither f nor g is injective.
• C. f is obviously injective.
• D. g is obviously injective.

Answer 1

The Correct Option is D

Given: Let \( S \), \( T \), and \( U \) be three non-empty sets, with functions \( f: S \to T \), \( g: T \to U \), and the composed function \( g \circ f: S \to U \). We are told that \( g \circ f \) is injective. We need to determine the correct property of \( f \) and \( g \). Step 1: Understanding injectivity of \( g \circ f \). A function is injective (or one-to-one) if distinct inputs map to distinct outputs. In other words, a function \( h \) is injective if for any \( x_1, x_2 \), we have: $$ h(x_1) = h(x_2) \implies x_1 = x_2. $$ Since \( g \circ f \) is injective, we know that: $$ g(f(x_1)) = g(f(x_2)) \implies x_1 = x_2. $$ Step 2: What does this imply about \( f \) and \( g \)? For the condition \( g(f(x_1)) = g(f(x_2)) \) to imply \( x_1 = x_2 \), the function \( f \) must map distinct elements of \( S \) to distinct elements of \( T \), and the function \( g \) must also map distinct elements of \( T \) to distinct elements of \( U \). This means that: - If \( f \) is not injective, it could map distinct elements in \( S \) to the same element in \( T \), which could still allow \( g \circ f \) to be injective if \( g \) differentiates between these elements in \( T \). Hence, \( f \) does not necessarily need to be injective. - However, if \( g \) is not injective, it could map different elements of \( T \) to the same element in \( U \), which would violate the injectivity of \( g \circ f \). Therefore, \( g \) must be injective. Conclusion: The function \( g \) must be injective for \( g \circ f \) to be injective. However, \( f \) does not necessarily have to be injective. Thus, the correct answer is: g is obviously injective.