Question

Consider functions \[f\] and \[g\] such that composite \[g \circ f\] is defined and is one are \[g\] both necessarily one-one?

Answer 1

Hint: Here in this question, we need to check whether both \[g \circ f\] and \[g\] is necessarily one-one or not. For this, first we need to consider a set of functions \[f\] and \[g\] and further define a composition function \[g \circ f\] then check the one-one condition to get the required solution.

Complete step by step answer:
Let’s consider a two functions \[f\left( x \right)\] and \[g\left( x \right)\], then function \[f\] ranges from \[f:A \to B\] and \[g\] ranges from \[g:B \to C\].
Now, the composite function \[g\] and \[f\] \[g \circ f:A\because C\], then defined as \[g \circ f:A \to C\] is one-one.
Now, we are to prove that \[f\] is one -one if possible. Suppose, that \[f\] is not one-one.
Let there exists a \[{x_1}\], \[{x_2}\] \[ \in A\] such that \[{x_1} \ne {x_2}\] But \[f\left( {{x_1}} \right) = f\left( {{x_2}} \right)\], then
\[ \Rightarrow \,\,\,\,g\left( {f\left( {{x_1}} \right)} \right) = g\left( {f\left( {{x_2}} \right)} \right)\]
Then,
\[ \Rightarrow \,\,\,\,g \circ f\left( {{x_1}} \right) = g \circ f\left( {{x_2}} \right)\]
Therefore, \[{x_1}\], \[{x_2}\] \[ \in A\] such that \[{x_1} \ne {x_2}\] but \[g \circ f\left( {{x_1}} \right) = g \circ f\left( {{x_2}} \right)\].
Therefore, \[g \circ f\] is not one which is against the given hypothesis that \[g\] of is one -one superposition is wrong.

Now, let \[f:\left\{ {1,2,3,4} \right\} \to \left\{ {1,2,3,4,5,6} \right\}\]

\[f\] is one-one and \[g:\left\{ {1,2,3,4,5,6} \right\} \to \left\{ {1,2,3,4,5,6} \right\}\]

\[g\] is not one-one. The composite function \[g \circ f\]

Hence, which shows that \[g \circ f\] is one-one. \[f\] and \[g\] are not necessarily one-one.

Note: One to one function basically denotes the mapping between the two sets. A function $g$ is one-to-one if every element of the range of g corresponds to exactly one element of the domain of $g$. A function \[f:A \to B\] is said to be an onto function if \[f\left( A \right)\], the image of A equal to B. that is \[f\] is onto if every element of B the co-domain is the image of at least one element of A the domain.