Question

Q : 5). If f is function from A to B and g is function B to C and both f and g are onto.
Show that go f is also onto. Is go f one-to-one if both f and g are one-to-one.


Q : 6) Let f, g and h: R → R be defined by (R is the set of real numbers)
f(x) = x + 2, g(x) = (1 + x2)
-1, h(x) = 3.
Compute f -1g(x) and hf (g f -1) (hf(x)).

Answer 1

Question #77239

1. If $f$ is function from $A$ to $B$ and $g$ is function $B$ to $C$ and both $f$ and $g$ are onto. Show that $g \circ f$ is also onto. Is $g \circ f$ one-to-one if both $f$ and $g$ are one-to-one.

Solution

Let choose an arbitrary $c \in C$, since $g: B \to C$ is onto map, then there exists $b \in B$ such that $g(b) = c$. Since $f: A \to B$ is onto map, there exists $a \in A$ such that $f(a) = b$. Therefore, $c = g(b) = g(f(a))$ and it shows that $g \circ f$ is onto map.

For every $a_1 \neq a_2$ from one-to-one property of $f$ we have $f(a_1) \neq f(a_2)$ and from one-to-one property of $g$ we have $g(f(a_1)) \neq g(f(a_2))$. So, if both $f$ and $g$ are one-to-one maps the map $g \circ f$ is also one-to-one.

2. Let $f, g$ and $h: \mathbb{R} \to \mathbb{R}$ be defined by ($\mathbb{R}$ is the set of real numbers) $f(x) = x + 2$, $g(x) = (1 + x^2)^{-1}$, $h(x) = 3$. Compute $f^{-1}g(x)$ and $hf(gf^{-1})(hf(x))$.

Solution

If $f(x) = x + 2$, then $f^{-1} = x - 2$ and $f^{-1}g(x) = f^{-1}((1 + x^2)^{-1}) = (1 + x^2)^{-1} - 2$.

$h(f(x)) = h(x + 2) = 3$

$g(f^{-1})(3) = g(3 - 2) = g(1) = 2^{-1} = 0.5$

$hf(0.5) = h(0.5 + 2) = h(2.5) = 3$, so

$hf(gf^{-1})(hf(x)) = 3.$