Give examples of function:
`f:NrarrN and g:NrarrN`
such that gof is onto but f is not onto.

Answer

To solve the problem, we need to provide examples of two functions \( f: \mathbb{N} \to \mathbb{N} \) and \( g: \mathbb{N} \to \mathbb{N} \) such that the composition \( g \circ f \) is onto, but \( f \) itself is not onto.

### Step 1: Define the function \( f \)

Let's define the function \( f \) as follows:
\[ f(x) = x + 1 \]
This function takes a natural number \( x \) and maps it to \( x + 1 \).

### Step 2: Show that \( f \) is not onto

To prove that \( f \) is not onto, we need to show that there exists at least one natural number \( y \) in the codomain that does not have a pre-image in the domain.

- The codomain of \( f \) is \( \mathbb{N} \), which includes all natural numbers \( 1, 2, 3, \ldots \).
- For \( f(x) = y \), we have \( x + 1 = y \).
- Rearranging gives \( x = y - 1 \).

Now, if we take \( y = 1 \):
\[ x = 1 - 1 = 0 \]
Since \( 0 \) is not a natural number, there is no \( x \in \mathbb{N} \) such that \( f(x) = 1 \). Thus, \( f \) is not onto.

### Step 3: Define the function \( g \)

Now, let's define the function \( g \) as:
\[ g(x) = \begin{cases} 1 & \text{if } x = 1 \\ x - 1 & \text{if } x > 1 \end{cases} \]
This function maps \( 1 \) to \( 1 \) and any natural number greater than \( 1 \) to its predecessor.

### Step 4: Show that \( g \circ f \) is onto

Now, we need to show that the composition \( g \circ f \) is onto.

1. We calculate \( g(f(x)) \):
\[ g(f(x)) = g(x + 1) \]
2. For \( x = 1 \):
\[ g(f(1)) = g(1 + 1) = g(2) = 2 - 1 = 1 \]
3. For \( x > 1 \):
\[ g(f(x)) = g(x + 1) = (x + 1) - 1 = x \]

Now, let's analyze the outputs:
- For \( x = 1 \), \( g(f(1)) = 1 \).
- For \( x = 2 \), \( g(f(2)) = g(3) = 3 - 1 = 2 \).
- For \( x = 3 \), \( g(f(3)) = g(4) = 4 - 1 = 3 \).
- And so on...

Thus, for every natural number \( y \), we can find an \( x \) such that \( g(f(x)) = y \). Therefore, \( g \circ f \) is onto.

### Conclusion

We have defined:
- \( f(x) = x + 1 \) (not onto)
- \( g(x) = 1 \) if \( x = 1 \) and \( g(x) = x - 1 \) if \( x > 1 \) (such that \( g \circ f \) is onto).