Question

Show that if $f:A \to B$ and $g:B \to C$ are onto, then $fog:A \to C$ is also onto.

Answer 1

To solve this we should remember the concept of function onto functions. To show this first we take the if part of the problem and we show the then part of the given problem.

Complete Step by Step Solution:
About functions : Let A and B be two non empty sets. A function f from A to B, denoted by $f:A \to B$ is a rule that assigns each member of A an unique member of B. here A is called the domain off while B is called the co domain of f. the set of all those members of B that are assigned by the rule f to some member of A is called the range of f.
Onto function : The onto function is also called as surjection defined as a function $f:A \to B$ if $f\left( A \right)$ , the image of A equals 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.
$f:A \to B$ is onto $\Leftrightarrow$ for every $y \in B$ there is at least one $x \in A$ such that $f\left( x \right) = y$ if and only if $f\left( A \right) = B$.
The objective of the problem is to show that if $f:A \to B$ and $g:B \to C$ are onto, then $fog:A \to C$ is also onto.
Since $g:B \to C$ is onto
Suppose $t \in C$ , then there exists a pre image in B.
Let that preimage be y.
Hence, $y \in B$ such that $g\left( y \right) = t$
Since , $f:A \to B$ is onto
If $y \in B$ then there exists a pre image in A
Let that preimage be x.
Hence , $x \in A$ such that $f\left( x \right) = y$
Now we find $\left( {gof} \right)\left( x \right)$
$\left( {gof} \right)\left( x \right) = g\left( {f\left( x \right)} \right)$
We have that $f\left( x \right) = y$
On substituting we get
$\left( {gof} \right)\left( x \right) = g\left( y \right)$
Here we have $g\left( y \right) = t$
$\left( {gof} \right)\left( x \right) = g\left( y \right) = t$
For every x in A , there is an image T in C.
Thus $gof$ is also onto.

Note:
In the product function$gof$ the co domain of f is the domain of g. the domain of $gof$ is the domain of f. The co domain of $gof$ and g is the same set.