Short Answer

Suppose that g is a function from A to B and f is a function from B to C.
Show that if both f and g are one-to-one functions, then $f \circ g$ is also one-to-one.
Show that if both f and g are onto functions, then $f \circ g$ is also onto.

$f \circ g$ is onto.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step: 1

a. Let x and y be distinct elements of A. Because g is one-to-one, g(x) and g(y) are distinct elements of B. Because f is one-to-one, $f (g (x)) = (f \circ g) (x)$ and $f (g (y)) = (f \circ g) (y)$ are distinct elements of C.

Hence $f \circ g$ is one-to-one.

Step: 2

b. Let $y \in C$ .

Because f is onto $y = f (b)$ , for some $b \in B$ .

Now because g is onto $b = g (x)$ , for some $x \in A$ .

hence, $y = f (b) = f (g (x)) = (f \circ g) (x)$

If follows that $f \circ g$ is onto.

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Suppose that g is a function from A to B and f is a function from B to C.
Show that if both f and g are one-to-one functions, then $f \circ g$ is also one-to-one.
Show that if both f and g are onto functions, then $f \circ g$ is also onto.

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Suppose that g is a function from A to B and f is a function from B to C.
Show that if both f and g are one-to-one functions, then $f \circ g$ is also one-to-one.
Show that if both f and g are onto functions, then $f \circ g$ is also onto.