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One of the implications of surjection is that all elements of co-domain is related. It reduces the co-domain to range set. In other words, co-domain is also the range of the function.
$$\text{Co-domain of "f"}=\text{Range of "f"}$$
This equality of sets form one of the condition for testing a function to be surjection. Alternatively, we can check surjectivity by evaluating the rule of the function for the argument “x”. If the expression of “x” is valid for elements in co-domain, then the function is a surjection.
Problem 2 : Consider a function defined as :
$$f:R\to R\phantom{\rule{1em}{0ex}}by\phantom{\rule{1em}{0ex}}f\left(x\right)={x}^{3}+1\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in R$$
Determine whether the function is a surjection?
Solution : We solve the rule for argument “x” as :
$$y={x}^{3}+1$$
$$\Rightarrow x={\left(y-1\right)}^{1/3}$$
We see that expression on the right hand side is a valid real expression for all values of “y” in "R" i.e co-domain. Hence, given function is an onto function or surjection.
We have discussed in the beigining of this module that there is possibility that some of the elements of co-domains are not related. In that case, function is said to be into function.
One of the implications is that all elements of co-domain are not related to elements in domain set. In other words, range of the function is subset of the co-domain :
$$\text{Range of "f"}\subset \text{Co-domain of "f"}$$
We can check whether a given function is an into function or not by checking whether the function is an onto set or not. If the function is not an onto function, then it an into function. Alternatively, we can check the equality of co-domain and range set. If they are not equal, then the function is into function.
Problem 3 : Consider a function defined as :
$$f:R\to R\phantom{\rule{1em}{0ex}}by\phantom{\rule{1em}{0ex}}f\left(x\right)={x}^{2}+1\phantom{\rule{1em}{0ex}}forall\phantom{\rule{1em}{0ex}}x\in R$$
Determine whether the function is an into function?
Solution : The rule of the function is :
$$y={x}^{2}+1$$
The square of a real number is a positive number for all real number. Hence,
$$\Rightarrow y={x}^{2}+1>0$$
It means that images are only the right half of the real number i.e. from zero to infinity. But, the co-domain of the function is “R”. It means that left half of the co-domain i.e. from negative infinity to less than zero has no image in “A”. Therefore, the given function is an into function.
The bijection presents the most stringent condition for a function. Every element of both domain and co-domain is related to distinct element. This requirement is fulfilled when a function is both an injection and surjection.
The injection means that every element of domain is related to a distinct element in co-domain. On the other hand, surjection means that every element of co-domain is related, both distinctly and commonly. When conditions of injection and surjection are taken together, then it is also ensured that elements of co-domains are also related to distinct elements only.
Problem 4 : Consider a function defined as :
$$f:A\to B\phantom{\rule{1em}{0ex}}by\phantom{\rule{1em}{0ex}}f\left(x\right)=\frac{x-2}{x-3}$$
Determine domain (A) and co-domain(B) of the function so that it is a bijection.
Solution : For determining domain of the function, we inspect the given rule,
$$f\left(x\right)=\frac{x-2}{x-3}$$
We observe that the given rational function is defined for all values of “x” except for x = 3. Hence, its domain is :
$$\mathrm{Domain}=R-\left\{3\right\}$$
In order that the given function is a bijection, it should be both an injection and a surjection. For infectivity, we put :
$$f\left(x\right)=f\left(y\right)$$
$$\Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}$$
$$\Rightarrow \left(x-2\right)\left(y-3\right)=\left(x-3\right)\left(y-2\right)$$
$$\Rightarrow xy-3x-2y+6=xy-2x-3y+6$$
$$\Rightarrow -3x-2y=-2x-3y$$
$$\Rightarrow x=y$$
Hence, function is an injection for the domain as determined above. Now, for surjection we solve the rule of the function for the argument (x),
$$\Rightarrow y=\frac{x-2}{x-3}$$
$$\Rightarrow xy-3y=x-2$$
$$\Rightarrow x\left(y-1\right)=3y-2$$
$$\Rightarrow x=\frac{3y-2}{y-1}$$
This equation is valid for all real values of “y” except “1”. Hence, function is surjection for all real values of “y” except for “1”. Hence, co-domain for the function to be a surjection is :
$$\mathrm{Co-domain}=R-\left\{1\right\}$$
Thus, we conclude that the given function is bijection for the domain and co-domain as determined above.
$$\mathrm{Domain}=R-\left\{3\right\}$$
$$\mathrm{Co-domain}=R-\left\{1\right\}$$
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