Question

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

a) Show that if both f and g are one-to-one functions, then $\mathit{f}\mathbf{\circ }\mathit{g}$is also one-to-one. b) Show that if both f and g are onto functions, then $\mathit{f}\mathbf{\circ }\mathit{g}$is also onto.

Short answer

Answer:

\(f \circ g\) is onto.

Step-by-step solution

Step: 1

a) Let x and y be distinct elements of A. Because g is one-to-one, $g\left(x\right)$and $g\left(y\right)$are distinct elements of B. Because f is one-to-one, role="math" localid="1668411603503" $f\left(g\right(x\left)\right)=(f\circ g)\left(x\right)$and$f\left(g\right(y\left)\right)=(f\circ g)\left(y\right)$are distinct elements of C.

Hence $\text{(f∘g)}$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$(f\circ g)$is onto.