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Q33E

Expert-verifiedFound in: Page 154

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.

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

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

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

If follows that $\mathrm{f}\circ \mathrm{g}$ is onto.

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