If f: A rightarrow B and g: B rightarrow C ar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: $f: A \rightarrow B$ and $g: B \rightarrow C$. Also,$g \circ f: A \rightarrow C$ is a bijection. First,we prove $f$ is an injection: Let $x

Vedclass · Accepted Answer

$f$ is an injection and $g$ is a surjection

Vedclass · Answer

(B) Given: $f: A \rightarrow B$ and $g: B \rightarrow C$. Also,$g \circ f: A \rightarrow C$ is a bijection. First,we prove $f$ is an injection: Let $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$. Then $g(f(x_1)) = g(f(x_2))$,which implies $(g \circ f)(x_1) = (g \circ f)(x_2)$. Since $g \circ f$ is a bijection,it is injective,so $x_1 = x_2$. Thus,$f$ is an injection. Next,we prove $g$ is a surjection: Let $z \in C$. Since $g \circ f: A \rightarrow C$ is a bijection,it is surjective. Therefore,there exists $x \in A$ such that $(g \circ f)(x) = z$. This means $g(f(x)) = z$. Since $f(x) \in B$,let $y = f(x)$. Then $g(y) = z$ for some $y \in B$. Thus,$g$ is a surjection. Therefore,$f$ is an injection and $g$ is a surjection.

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are two functions such that $g \circ f: A \rightarrow C$ is a bijection,then which one of the following is always true?

Similar Questions

Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x \in R$. Then the set of all $x$ satisfying $(f \circ g \circ g \circ f)(x)=(g \circ g \circ f)(x)$,where $(f \circ g)(x)=f(g(x))$,is

If $f(x)=\sqrt{x}$ $(x \geq 0)$ and $g(x)=1+x^2$, then $(f \circ g)^{\prime}(1)=$

If $f: R \rightarrow R$ and $g: R^{+} \rightarrow R$ are such that $g\{f(x)\}=|\sin x|$ and $f\{g(x)\}=(\sin \sqrt{x})^2$, then a possible choice for $f$ and $g$ is

If $f(x) = x^2 - 1$ and $g(x) = 3x + 1$,then $(gof)(x) = $

Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ be functions such that $g \circ f: S \rightarrow U$ is surjective. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module