Answer
No inverse
Work Step by Step
Given the function
\[
h: 2\mathbb{Z} \;\to\; 2\mathbb{Z}
\quad\text{defined by}\quad
h(n) \;=\; 2n \quad \text{for } n \in 2\mathbb{Z}.
\]
Recall that \(2\mathbb{Z}\) denotes the set of all even integers, i.e., \(\{\,2k : k \in \mathbb{Z}\}\).
---
## 1. Injectivity
To test injectivity, suppose \(h(n_1) = h(n_2)\). Then
\[
2n_1 \;=\; 2n_2 \;\;\Longrightarrow\;\; n_1 \;=\; n_2.
\]
Thus \(h\) is injective on \(2\mathbb{Z}\).
---
## 2. Surjectivity
We must see if for **every** \(y \in 2\mathbb{Z}\), there is some \(n \in 2\mathbb{Z}\) such that \(h(n) = y\).
- A generic element of \(2\mathbb{Z}\) has the form \(y = 2m\) for some integer \(m\).
- We want \(n \in 2\mathbb{Z}\) (so \(n = 2k\) for some \(k\in\mathbb{Z}\)) such that
\[
h(n) \;=\; 2n \;=\; 2(2k) \;=\; 4k \;=\; 2m.
\]
This simplifies to \(4k = 2m \Rightarrow 2k = m \Rightarrow k = m/2\).
However, \(k\) must be an integer. That forces \(m\) to be even. In other words, \(y\) must be a multiple of 4 for \(y\) to lie in the image of \(h\). Thus:
- If \(m\) is even, \(m = 2k\), then \(y = 2m\) is indeed a multiple of 4, and there is an \(n \in 2\mathbb{Z}\) mapping to \(y\).
- If \(m\) is odd, \(y = 2(2k+1)\) is a multiple of 2 but **not** a multiple of 4, and there is **no** \(n \in 2\mathbb{Z}\) such that \(h(n) = y\).
Hence \(h\) does **not** hit all even integers—only the multiples of 4. It fails to be onto \(2\mathbb{Z}\).
---
## 3. Conclusion
Since \(h\) is injective but **not** surjective as a map \(2\mathbb{Z}\to 2\mathbb{Z}\), it is **not** a one-to-one correspondence. Consequently, it has **no inverse function** on \(2\mathbb{Z}\).
