Yes, your representation is correct according to the solution. Let's break down your diagrams for clarity:

Diagram Analysis

First Diagram (Top)

This diagram shows the case where $n_1$ and $n_2$ are both odd or both even:

• The left set (a) represents possible values for $n$, which can be 1 or -1.
• The right set (b) also represents the same possible values for $f(n)$, which can be 1 or -1.
• If $n_1$ and $n_2$ are both even or both odd, $f(n_1)$ and $f(n_2)$ map to the same point in $\mathbb{Z}$, ensuring $n_1 = n_2$.

Second Diagram (Bottom)

This diagram shows the case where one of $n_1$ or $n_2$ is odd and the other is even:

• The left set (a) represents possible values for $n$, which can be 1 or -1.
• The right set (b) also represents the same possible values for $f(n)$, which can be 1 or -1.
• If $n_1$ is odd and $n_2$ is even, $f(n_1)$ maps to one point (say, 1) and $f(n_2)$ maps to another point (say, -1), ensuring that $n_1 \neq n_2$.

Your drawings correctly illustrate the cases discussed in the solution:

1. When $n_1$ and $n_2$ are both odd or both even, they map to the same point.
2. When $n_1$ is odd and $n_2$ is even (or vice versa), they map to different points.

Therefore, your visual representation aligns with the proof and accurately depicts the injectivity of the function $f(n) = (-1)^n \cdot n$.

Do you want details or have any questions? Here are some more questions you might find interesting:

1. How do you prove that a function is surjective?
2. What is the difference between injective, surjective, and bijective functions?
3. How can you determine if a function is bijective?
4. What are some examples of non-injective functions?
5. How do you prove the injectivity of functions involving trigonometric identities?

Tip: Visual aids like diagrams can be very helpful in understanding and proving properties of functions, especially injectivity and surjectivity.