Surjective functions

Say we have the function below,

\[f:\mathbb{R}\mapsto \mathbb{R}\]

\[f(x)=3x\]

The domain of this function is the set of all real numbers.

The codomain of this function is the set of all real numbers.

Is this a surjective function?

Solution

In order to test if this function is surjective, we need to check whether the range and the codomain of the function \(f\) are the same.

Here the codomain is the set of real numbers as stated in the question.

Now, in order to determine the range, we should think of all the possible outcomes of the function into consideration. Taking into account that the inputs are the set of all real numbers, multiplying each of them by 3 to produce the set of outcomes, which is nothing but the range, will lead us also to the set of the real numbers.

Thus, the range and the codomain of the function are the same and hence the function is surjective.

Suppose we are given two surjective functions \(f\) and \(g\) where

\[f:\mathbb{R}\mapsto \mathbb{R} \quad\text{and}\quad g:\mathbb{R}\mapsto \mathbb{R}\]

The function \(f\) is defined by

\[f(x)=3x\]

The function \(g\) is defined by

\[g(x)=2x\]

Does the composition \(g\circ f\) yield a surjective function?

Solution

Since \(f:\mathbb{R}\mapsto\mathbb{R}\) and \(g:\mathbb{R}\mapsto\mathbb{R}\), then \(g\circ f:\mathbb{R}\mapsto\mathbb{R}\).

Let us consider an arbitrary element, \(z\) in the codomain of \(g\circ f\), our aim is to prove that for every \(z\) in the codomain of \(g\circ f\) there exists one element \(x\) in the domain of \(g\circ f\) such that \(z=g\circ f(x)=g(3x)=2(3x)=6x\).

Since \(g\) is surjective, there exists some arbitrary element \(y\) in \(\mathbb{R}\) such that \(g(y)=z\) but \(g(y)=2y\), thus \(z=g(y)=2y\).

Similarly, since \(f\) is surjective, there exists some arbitrary element \(x\) in \(\mathbb{R}\) such that

\[f(x)=y\]

but \(f(x)=3x\), thus \(y=f(x)=3x\).

Therefore, we have \(z=g(y)=2y=2(3x)=6x\).

We deduce thus that \(g\circ f\) is surjective.

Given the function \(f\) where \(f:\mathbb{Z}\mapsto \mathbb{Z}\) defined over the set of integers, \(\mathbb{Z}\), where

\[f(x)=x+4\]

show whether this function is surjective or not.

Solution

We shall first claim that this function is surjective. We now need to show that for every integer \(y\), there exists an integer \(x\) such that \(f(x) = y\).

Taking our equation as

\[f(x)=y \Rightarrow y=x+4\]

We shall now work backward toward our goal by solving for \(x\). Assume that for any element \(y\in\mathbb{Z}\) there exists an element \(x\in\mathbb{Z}\) such that

\[x=y-4\]

This is done by rearranging the previous equation so that \(x\) becomes the subject. Then, by this choice of \(x\) and by the definition of \(f(x)\), we obtain

\[\begin{align}f(x)&=f(y-4)\\ \Rightarrow f(x)&=(y-4)+4\\ \Rightarrow f(x)&=y\end{align}\]

Hence, \(y\) is an output of \(f\) which indicates that \(f\) is indeed surjective.

Say we are given the exponential function, \(f:\mathbb{R}\mapsto\mathbb{R}\) defined by

\[f(x)=e^x\]

Note that \(\mathbb{R}\) represents the set of real numbers. The graph of this function is shown below.

Fig. 2. Exponential graph.

By observing this graph, determine whether the function surjective or not.

Solution

Here, the codomain is the set of real numbers as given in the question.

Referring to the graph, the range of this function is only defined over the set of positive real numbers including zero. In other words, the range of \(f\) is \(y\in [0,\infty)\). Since the codomain of \(f\) is not equal to the range of \(f\), we can conclude that \(f\) is not surjective.

Say we are given the standard cubic function, \(g:\mathbb{R}\mapsto\mathbb{R}\) defined by

\[g(x)=x^3\]

The graph of this function is shown below.

Fig. 3. Standard cubic graph.

By observing this graph, determine whether the function surjective or not.

Solution

In this case, the codomain is the set of real numbers as given in the question.

Looking at the graph, notice that the range of this function is also defined over the set of real numbers. This means that the range of \(g\) is \(y\in\mathbb{R}\). As the codomain of \(g\) is equal to the range of \(g\), we can infer that \(g\) is surjective.

Using the horizontal line test, determine whether the graph below is surjective or not. The domain and range of this graph is the set of real numbers.

Fig. 4. Example A.

Solution

Let us construct three horizontal lines on the graph above, namely \(y=-1\), \(y=0.5\) and \(y=1.5\). This is shown below.

Fig. 5. Solution to Example A.

Now looking at the intersecting points on this graph, we observe at \(y=1.5\), the horizontal line intersects the graph once. At \(y=-1\) and \(y=0.5\), the horizontal line intersects the graph three times. In all three instances, the horizontal line intersects the graph at least once. Thus, the graph satisfies the condition for a function to be surjective.

As before, apply the horizontal line test to decide if the following graph is surjective or not. The domain and range of this graph is the set of real numbers.

Fig. 6. Example B.

Solution

As before, we shall construct three horizontal lines on the graph above, namely \(y=-5\), \(y=-2\) and \(y=1\). This is shown below.

Fig. 7. Solution to Example B.

Notice how at \(y=-5\) and \(y=1\) the horizontal line intersects the graph at one point. However, at \(y=-2\), the horizontal line test does not intersect the graph at all. Thus, the horizontal line test fails and is not surjective.

Consider the standard square function, \(f:\mathbb{R}\mapsto\mathbb{R}\) defined by

\[f(x)=x^2\]

Check whether the function is surjective or not.

Solution

Let us sketch this graph.

Fig. 8. Standard square graph.

Here, the codomain is the set of real numbers as given in the question.

Referring to the sketch above, the range of this function is only defined over the set of positive real numbers including zero. Thus, the range of \(f\) is \(y\in [0,\infty)\). However, the codomain includes all negative real numbers as well. Since the codomain of \(f\) is not equal to the range of \(f\), we can conclude that \(f\) is not surjective.

Suppose we have two sets, \(P\) and \(Q\) defined by \(P =\{3, 7, 11\}\) and \(Q = \{2, 9\}\). Suppose we have a function \(g\) such that

\[g = \{(3, 2), (7, 2), (11, 9)\}\]

Verify that this function is surjective from \(P\) to \(Q\).

Solution

The domain of set \(P\) is equal to \(\{3, 7, 11\}\). From our given function, we see that each element of set \(P\) is assigned to an element such that both \(3\) and \(7\) share the same image of \(2\) and \(11\) has an image of \(9\). This means that the range of the function is \(\{2, 9\}\).

Since the codomain \(Q\) is equal to \(\{2, 9\}\) as well, we find that the range of the function is also equal to set \(Q\). Thus, \(g:P\mapsto Q\) is a surjective function.

Given the function \(h:\mathbb{R}\mapsto\mathbb{R}\) defined by,

\[h(x)=2x-7\]

Check whether this function is surjective or not.

Solution

We shall first assume that this function is surjective. Our goal is to show that for every integer \(y\), there exists an integer \(x\) such that \(h(x) = y\).

Taking our equation as

\[h(x)=y\]

\[\Rightarrow 2x-7\]

We shall now work backward toward our goal by solving for \(x\). Suppose that for any element \(y\in \mathbb{R}\) there exists an element \(x\in\mathbb{R}\) such that

\[x=\dfrac{y+7}{2}\]

This is done by rearranging the previous equation so that \(x\) becomes the subject as below.

\[\begin{align}y&=2x-7\\ \Rightarrow 2x&=y+7\\ \Rightarrow x&=\dfrac{y+7}{2}\end{align}\]

Then, by this choice of \(x\) and by the definition of \(h(x)\), we obtain

\[\begin{align} h(x)&=h\left(\dfrac{y+7}{2}\right)\\ \Rightarrow h(x)&=\cancel{2}\left(\dfrac{y+7}{\cancel{2}}\right)-7\\ \Rightarrow h(x)&=y+7-7\\ \Rightarrow h(x)&=y \end{align}\]

Hence, \(y\) is an output of \(h\) which indicates that \(h\) is indeed surjective.