Question

Let $f: A \rightarrow B$ and $g: C \rightarrow A$ be functions. Then which of the following options is/are correct?

• A. If $f$ and $g$ are Injective, then $f \circ g$ is also Injective.
• B. If $f$ and $g$ are Surjective, then $f \circ g$ is also Surjective.
• C. If $f \circ g$ is Injective, then $g$ MUST be Injective.
• D. If $f \circ g$ is Injective, then $f$ need not be Injective.

Answer 1

Answer: (A), (B), (C), (D)

(A), (B), (C), (D)

Answer 2

• (A) If $f$ and $g$ are Injective, then $f \circ g$ is also Injective.
  Let $x_1, x_2 \in C$ such that $(f \circ g)(x_1) = (f \circ g)(x_2)$. This means $f(g(x_1)) = f(g(x_2))$. Since $f$ is injective, $g(x_1) = g(x_2)$. Since $g$ is injective, $x_1 = x_2$. Thus, $f \circ g$ is injective.
  * (B) If $f$ and $g$ are Surjective, then $f \circ g$ is also Surjective.
  Let $y \in B$. Since $f$ is surjective, there exists $a \in A$ such that $f(a) = y$. Since $g$ is surjective, there exists $c \in C$ such that $g(c) = a$. Then $(f \circ g)(c) = f(g(c)) = f(a) = y$. Thus, $f \circ g$ is surjective.
  * (C) If $f \circ g$ is Injective, then $g$ MUST be Injective.
  Let $x_1, x_2 \in C$ such that $g(x_1) = g(x_2)$. Applying $f$ to both sides, $f(g(x_1)) = f(g(x_2))$, which means $(f \circ g)(x_1) = (f \circ g)(x_2)$. Since $f \circ g$ is injective, $x_1 = x_2$. Therefore, $g$ is injective.
  * (D) If $f \circ g$ is Injective, then $f$ need not be Injective.
  Counterexample: Let $A = \{1, 2\}$, $B = \{3\}$, $C = \{4\}$. Define $f(1) = 3, f(2) = 3$ (not injective). Define $g(4) = 1$. Then $(f \circ g)(4) = f(g(4)) = f(1) = 3$. The function $f \circ g$ has a singleton domain, so it is injective, while $f$ is not.