Let $ f: \frac{\mathbb{Z}}{42\mathbb{Z}}\rightarrow \frac{\mathbb{Z}}{3\mathbb{Z}} × \frac{\mathbb{Z}}{7\mathbb{Z}} $
Such that $ f( z+ 42\mathbb{Z}) =(z+3\mathbb{Z}, z+7\mathbb{Z} )$
Find $f^{-1} ( ( 1+3\mathbb{Z}, -1+ 7\mathbb{Z} ) )$
And tell if the function is injective and if surjective.
Injective proof:
Let $(x+42\mathbb{Z})= (z+42\mathbb{Z})$
Therefore $ f (x+42\mathbb{Z})= (x+3\mathbb{Z},x+7\mathbb{Z})= (z+3\mathbb{Z},z+7\mathbb{Z})= f (z+42\mathbb{Z})$
Proving that the function is Indeed injective, but we also can see from cardinalities that the Cardinality of the domain is 42 and the codomain one is 21, and 42>21; therefore, the function is not injective ( what have I done wrong in the injective proof?)
Surjective: we already know that the function is surjective, but I want to "verify" it using the definition as an exercise. We need to find that for all elements of the codomain, their inverse image is not the empty set, let $ (a+3\mathbb{Z},b+7\mathbb{Z})$ be an element of the codomain, therefore, to find its inverse, we need to solve the system.
$x \equiv a \quad mod 3, x \equiv b \quad mod 7 $ that has solutions due to the Chinese remainder theorem, but how do we find the solution of the system? And how do we convert them as the inverse image?
To find the inverse of $ ( 1+3\mathbb{Z}, -1+ 7\mathbb{Z})$
We need to solve the associated system whose solutions are $x=13+ 21k$ after calculations. Having found the solutions, how do we find the inverse? I was told that the solution is $(13 +42\mathbb{Z}, 34+42\mathbb{Z})$
But how do I find it from the system's solution?
What you've done in your attempted proof of injectivity is show that if $\ x+42\mathbb{Z}\ $ and $\ z+42\mathbb{Z}\ $ are two synonymous expressions for the same element of $\ \mathbb{Z}/42\mathbb{Z}\ $, then $\ (x+3\mathbb{Z},x+7\mathbb{Z})\ $ and $\ (z+3\mathbb{Z},z+7\mathbb{Z})\ $ are synonymous expressions for the same element of $\ \mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/7\mathbb{Z}\ .$ That merely shows that the function $\ f\ $ is well-defined, not that it's injective. To show injectivity, you have to go back the other way—that is, show that if $\ f\big(x+42\mathbb{Z}\big)=f\big(z+42\mathbb{Z}\big)\ $ then $\ x+42\mathbb{Z}=z+42\mathbb{Z}\ .$ As you've already realised, you can't do this because $\ f\ $ isn't in fact injective. Your argument from cardinality is already sufficient to prove this. However, if $\ x-z\ $ is divisible by $\ 3\ $ and $\ 7\ $ but not by $\ 42\ $ then you'll have $\ x+42\mathbb{Z}\ne z+42\mathbb{Z}\ $ but $\ f\big(x+42\mathbb{Z}\big)=f\big(z+42\mathbb{Z}\big)\ ,$ which would also show that $\ f\ $ isn't injective, so you could also prove this by finding such a pair of numbers.
For the surjectivity, you know that if $\ x\ $ is any solution of the pair of congruences \begin{align} x&\equiv a\pmod{3}\label{e1}\tag{1}\\ x&\equiv b\pmod{7}\label{e2}\tag{2}\\ \end{align} then $\ f\big(x+42\mathbb{Z}\big)=(a+3\mathbb{Z}, b+3\mathbb{Z})\ .$ Since $\ 4\ $ and $\ 7\ $ are relatively prime, the Chinese remainder theorem tells you that the set of solutions of the pair of congruences (\ref{e1}) and (\ref{e2}) is $\ \{\,x+21k\,|\,k\in\mathbb{Z}\,\}\ $ where $\ x\ $ is the unique solution satisfying $\ x\in\{0,1,\dots,20\}\ .$ Any one of these solutions is sufficient to show that $\ (a+3\mathbb{Z}, b+3\mathbb{Z})\ $ lies in the range of $\ f\ .$ In the case when $\ a=1,b=-1\ $, you've correctly deduced that $\ 13+21k\ $ are solutions for any value of $\ k\in\mathbb{Z}\ .$ This means that $$ f\big(13+21k+42\mathbb{Z}\big)=\big(1+3\mathbb{Z},-1+7\mathbb{Z}\big) $$ for any value of $\ k\ .$ Since $\ 13+21k+42\mathbb{Z}=13+42\mathbb{Z}\ $ for any even value of $\ k\ $ and $\ 13+21k+42\mathbb{Z}=34+42\mathbb{Z}\ $ for any odd value of $\ k\ ,$ however, there are only two distinct pre-images of $\ \big(1+3\mathbb{Z},-1+7\mathbb{Z}\big)\ $—namely, $\ 13+42\mathbb{Z}\ $ and $\ 34+42\mathbb{Z}\ .$