In the previous section,
we saw composition of functions. We saw some solved examples also. In
this section, we will see a few more solved examples. Those solved examples will help us to learn the properties of composition of functions.
Solved example 17.18
Show that if f: A → B and g: B → C are one-one, then g∘f: A → C is also one-one
Solution:
1. In an one-one function, if (g∘f)(x1) is to be equal to (g∘f)(x2), then x1 must be equal to x2. This
condition can be used to prove that, a given function is an one-one function.
2. In our present case, suppose that, (g∘f)(x1) is equal to (g∘f)(x2). Then we can write:
$\begin{array}{ll}{} &{(g \circ f)(x_1)} & {~=~} &{(g \circ f)(x_2)} &{} \\
{\implies} &{g(f(x_1))} & {~=~} &{g(f(x_1))} &{} \\
{\implies} &{f(x_1)} & {~=~} &{f(x_2)~\color{magenta}{\text{- - - (I)}}} &{} \\
{\implies} &{x_1} & {~=~} &{x_2~\color{magenta}{\text{- - - (II)}}} &{} \\
\end{array}$
◼ Remarks:
•
Line marked as I:
We get this result because, g is one-one.
•
Line marked as II:
We get this result because, f is one-one.
Solved example 17.19
Show that if f: A → B and g: B → C are onto, then g∘f: A → C is also onto
Solution:
1. Given that g is onto. So for every element z of C, there will be a pre-image y in B.
♦ We can write: g(y) = z
2. Given that f is onto. So for every element y of B, there will be a pre-image x in A.
♦ We can write: f(x) = y
3. So we get: (f∘g)(x) = g(f(x)) = g(y) = z
• Taking the first and last items, we get: (f∘g)(x) = z
• That means, for every element z of C, there is a pre-image in A.
• Therefore, (g∘f) is an onto function
Solved example 17.20
Consider functions f and g such that composite g∘f is defined and is one-one. Are f and g both necessarily one-one.
Solution:
We will check this using an example. It can be written in 6 steps:
1. We have two sets:
• A = {1,2,3,4}
• B = {1,2,3,4,5,6}
2. We have two functions:
• f: A→B is defined by: f(x) = x
• g: B→B is defined by:
♦ g(x) = x, for x = 1,2,3,4
♦ g(5) = 5
♦ g(6) = 5
3. Based on the above steps, we can draw the Venn diagrams as shown in fig.17.9 below:
 |
| Fig.17.9 |
4. From the Venn diagram, we see that:
• (g∘f) is one-one.
• f is one-one.
• g is not one-one.
5. So we can write:
Even if the composite function (g∘f) is one-one, it is not necessary for both f and g to be one-one.
6. Let us see a comparison between the two solved examples:
(i) In a previous solved example 17.18, we see that:
If both f and g are one-one, then (g∘f) will be one-one.
(ii) In the present solved example 17.20, we see that:
If (g∘f) is one-one, then it is not necessary for both f and g to be one-one.
Solved example 17.21
Are f and g both necessarily onto, if (g∘f) is onto?
Solution:
We will check this using an example. It can be written in 6 steps:
1. We have two sets:
• A = {1,2,3,4}
• B = {1,2,3}
2. We have two functions:
• f: A→A is defined by:
♦ f(x) = x, for x = 1,2
♦ f(3) = 3
♦ f(4) = 3
• g: A→B is defined by:
♦ g(x) = x, for x = 1,2
♦ g(3) = 3
♦ g(4) = 3
3. Based on the above steps, we can draw the Venn diagrams as shown in fig.17.10 below:
 |
| Fig.17.10 |
• (g∘f) is onto.
• f is not onto.
• g is onto.
5. So we can write:
Even if the composite function (g∘f) is onto, it is not necessary for both f and g to be onto.
6. Let us see a comparison between the two solved examples:
(i) In a previous solved example 17.19, we see that:
If both f and g are onto, then (g∘f) will be onto.
(ii) In the present solved example 17.21, we see that:
If (g∘f) is onto, then it is not necessary for both f and g to be onto.
| • In general, it can be proved that: ♦ If (g∘f) is one-one, then f is one-one. ♦ If (g∘f) is onto, then g is onto. |
In the next section, we will see Invertible function.
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