Is $f: ( \mathbb{ Z}_{10},+) \rightarrow ( \mathbb{ Z}_{5} \times \mathbb{ Z}_{2},+), n\pmod {10} \mapsto (n \pmod 5,n \pmod 2) $ surjective?

Question

Is $f: ( \mathbb{ Z}_{10},+) \rightarrow ( \mathbb{ Z}_{5} \times \mathbb{ Z}_{2},+), n\pmod {10} \mapsto (n \pmod 5,n \pmod 2) $ surjective?

If I take an element in $\mathbb{ Z}_{5} \times \mathbb{ Z}_{2}$, say $(p,q)=f(n)$ I have to check this equation has always a solution for any value of p and q

so $([p]_5,[q]_2)=([n]_5,[n]_2)$

then I don't now how to continue. any help please?

Answer 1

So you want an $x \in \Bbb{Z}_{10}$ such that \begin{align*} x &\equiv a \pmod{5}\\ x &\equiv b \pmod{2} \end{align*} So $x=a+5k=b+2s$ for some $s,k \in \Bbb{Z}$. Thus $\boxed{5k-2s=b-a}$. But $\gcd(2,5)=1$ so we can express \begin{align*} 5(1)-2(2)&=1\\ 5(b-a)-2(2b-2a)&=b-a. \end{align*} So if we take $k=b-a$ (see boxed equation and the last equation above) we get $x=a+5(b-a)=\color{red}{5b-4a}$. You can verify that $$f(5b-4a)=([a]_{5},[b]_2)$$

In general use Chinese Remainder theorem.