is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

You will get a detailed answer to your question or assignment in the shortest time possible.

## Here's the Solution to this Question

Suppose $g : A\to B$ and $f : B\to C$  are functions.

a. Show that if $f\circ g$  is onto, then $f$  must also be onto.

Answer. Assume that $f\circ g$  is onto. Let $c\in C$ . Since $f\circ g$  is onto, there is $a\in A$  such that $c = (f\circ g)(a)$  by the definition of onto function. By the definition of function composition, $c = f(g(a))$ . Let $b = g(a)$ . This $b$  belongs to $B$  and is such that $c = f(b)$. Therefore, for every $c\in C$ , there is $b\in B$  such that $c = f(b)$ . By the definition of onto function, $f$  is onto.

b. Show that if $f\circ g$  is one-to-one, then $g$  must also be one-to-one.

Answer. Let $a_1, a_2\in A$  such that $g(a_1) = g(a_2)$ . Then

$(f\circ g)(a_1) = f(g(a_1)) = f(g(a_2))= (f\circ g)(a_2).$

Since $f\circ g$  is one-to-one, $a_1=a_2$ . Therefore, for all $a_1, a_2\in A$  such that $g(a_1) = g(a_2)$ , $a_1=a_2$ . By the definition of one-to-one function, $g$  is one-to-one.

c. Show that if $f\circ g$  is a bijection, then $g$  is onto if and only if $f$  is one-to-one.

Answer. Assume that $f\circ g$  is a bijection.

Assume that $g$  is onto. Let $b_1, b_2\in B$  such that $f(b_1) = f(b_2)$ . Since $g$  is onto, by the definition of onto function, there are $a_1, a_2\in A$  such that $b_1 = g(a_1)$  and $b_2 = g(a_2)$ . We have

$(f\circ g)(a_1) = f(g(a_1)) = f(b_1)$

$= f(b_2) = f(g(a_2)) = (f\circ g)(a_2).$

Since $f\circ g$  is a bijection, it is one-to-one, hence $a_1=a_2$ , and

$b_1 = g(a_1) = g(a_2) = b_2.$

•  Therefore, for all $b_1, b_2\in B$  such that $f(b_1) = f(b_2)$ , $b_1=b_2$ . By the definition of one-to-one function, $f$  is one-to-one.
• Assume that $f$  is one-to-one. Let $b\in B$ . Since $f\circ g$  is a bijection, it is onto, hence there is $a\in A$  such that $f(b) = (f\circ g)(a) = f(g(a))$ . Since $f$  is one-to-one, $b = g(a)$ . Therefore, for every $b\in B$ , there is $a\in A$  such that $b = g(a)$ . By the definition of onto function, $g$  is onto.