Question

Show that if $f{\rm{ }}:{\rm{ }}A \to B$ and $g{\rm{ }}:{\rm{ }}B \to C$ are one-one then $g \circ f{\rm{ }}:{\rm{ }}A \to C$ is also one-one.

Answer 1

If f(x) = f(y) implies x=y, then f is one-to-one mapped, or f is 1-1 is the way used to prove the given statement. Here using the given condition of function g and f we define the functional values and then using the definition of composition we will prove the result.

Complete step-by-step answer:
It is given that, $f{\rm{ }}:{\rm{ }}A \to B$ and $g{\rm{ }}:{\rm{ }}B \to C$ are one-one.
From the condition$f{\rm{ }}:{\rm{ }}A \to B$ is one-one, we have
$f({x_1}) = f({x_2}) \Rightarrow {x_1} = {x_2}$
From the condition$g{\rm{ }}:{\rm{ }}B \to C$ is one – one, we get,
$g({x_1}) = g({x_2}) \Rightarrow {x_1} = {x_2}$
“If $g \circ f{\rm{ }}:{\rm{ }}A \to C$ is also one-one then for $gof({x_1}) = gof({x_2})$ we will get,${x_1} = {x_2}$”
Let us consider $gof({x_1}) = gof({x_2})$
Then using the definition of composition we get,
$g(f({x_1})) = g(f({x_2}))$
Given that, $g{\rm{ }}:{\rm{ }}B \to C$ is one – one using the condition of one-one function, we get,
g(f({x_1})) = g(f({x_2}))$$$$ \Rightarrow f({x_1}) = f({x_2}).
Since$f{\rm{ }}:{\rm{ }}A \to B$ is one-one using the condition of one-one function, we get,
$f({x_1}) = f({x_2}) \Rightarrow {x_1} = {x_2}$
Thus from consideration we have got,
gof({x_1}) = gof({x_2})$$$$ \Rightarrow {x_1} = {x_2}.
Hence for the function$g \circ f{\rm{ }}:{\rm{ }}A \to C$ , if $gof({x_1}) = gof({x_2})$ then ${x_1} = {x_2}$.
Hence by definition of one-one we say that $g \circ f{\rm{ }}:{\rm{ }}A \to C$ is one-one.
Therefore, if $f{\rm{ }}:{\rm{ }}A \to B$ and $g{\rm{ }}:{\rm{ }}B \to C$ are one-one then $g \circ f{\rm{ }}:{\rm{ }}A \to C$ is also one-one.

Additional information:
One - one function basically denotes the mapping of two sets. A function g is one-one if every element of the range of g corresponds to exactly one element of the domain of g. One-to-one is also written as 1-1. A function $f()$ is a method, which relates elements/values of one variable to the elements/values of another variable, in such a way that the elements of the first variable identically determine the elements of the second variable.

It could be defined as each element of Set A has a unique element on Set B.
In brief, let us consider ‘f’ is a function whose domain is set A. The function is said to be injective if for all x and y in A,
Whenever$f\left( x \right) = f\left( y \right)$ , then $x = y$
And equivalently, if $x{\rm{ }} \ne {\rm{ }}y$ , then $f\left( x \right){\rm{ }} \ne {\rm{ }}f\left( y \right)$ .
Formally, it is stated as, if $f\left( x \right) = f\left( y \right)$ implies $x = y$, then f is one-to-one mapped, or f is 1-1.

Note: $g \circ f$ is the composition of function g and f which is defined as follows,
$g \circ f(x) = g(f(x))$ Here the function g is operated first and the function f is operated next.