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Expert-verified Found in: Page 154 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # 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 $\mathrm{f}\circ \mathrm{g}$ is also one-to-one.Show that if both f and g are onto functions, then $\mathrm{f}\circ \mathrm{g}$ is also onto.

$\mathrm{f}\circ \mathrm{g}$ is onto.

See the step by step solution

## 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, $f\left(g\left(x\right)\right)=\left(\mathrm{f}\circ \mathrm{g}\right)\left(\mathrm{x}\right)$ and $f\left(g\left(y\right)\right)=\left(\mathrm{f}\circ \mathrm{g}\right)\left(y\right)$ are distinct elements of C.

Hence $\mathrm{f}\circ \mathrm{g}$ is one-to-one.

## Step: 2

b. Let $y\in C$.

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

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

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

If follows that $\mathrm{f}\circ \mathrm{g}$ is onto. ### Want to see more solutions like these? 