$\varphi_{10} : \mathbb{N} \rightarrow \mathbb{N}, n\rightarrow \{6n+1 ($if n is even$), 6n-1$ if n is odd}
Well, I've calculated it, which gives that $1 \rightarrow 5$, $2 \rightarrow 13$, $3 \rightarrow 17$, $4 \rightarrow 25$, $5 \rightarrow 29$, $6 \rightarrow 37$, $7 \rightarrow 41...$
I think that this is surjective, because for $ \rho: A \rightarrow B, (\forall b \in B)(\exists a \in A)(a \rho = b) $ I assume that this is true.
And it's injective if $(\forall a_{1},a_{2} \in A)(a_{1} \rho = a_{2} \rho \rightarrow a_{1}=a_{2})$ so it has to be an injection but in which way can I show it exactly? And how can I get an inverse for it?
Hints:
For injectivity:
Consider two points $n_1\neq n_2$ and imply $\varphi_{10}(n_1)\neq \varphi_{10}(n_2)$. You have to consider three cases: First, $n_1$ and $n_2$ are even. Next, $n_1$ is even and $n_2$ is odd. Finally, $n_1$ and $n_2$ are odd.
For not surjectivity:
You can see that $\varphi_{10}(n)\geq 6n-1$ for all $n\in \mathbb N$.
For the partial inverse:
The first step is to understand the image $\varphi_{10}(\mathbb N)$. For an even number $n$, you find a number $k$ such that $n=2k$ and for an odd number $n$, you find a number $k$ such that $n=2k+1$. Hence, you should consider $\varphi_{10}(2k)$ and $\varphi_{10}(2k+1)$. Then you can see how to characterize a number $m\in \varphi_{10}(\mathbb N)$ depending on $k$ and find the partial inverse function.