Question

Suppose that g is a function from A to B and f is a function from B to C.

1. Show that if both f and g are one-to-one functions, then$\mathrm{f}\circ \mathrm{g}$is also one-to-one.
2. Show that if both f and g are onto functions, then $\mathrm{f}\circ \mathrm{g}$ is also onto.

Short answer

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

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, $f\left(g\right(x\left)\right)=(\mathrm{f}\circ \mathrm{g})(\mathrm{x})$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.

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

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