Proof of Injection and Surjection I am having trouble proving the function f is injective and surjective. $f$ is a function from $\mathbb{Z}\times{Z} \to $\mathbb{Z}\times{Z}$ and $f(x,y) = (5x-y,x+y)$.
I know it should be fairly easy but am still having trouble with the basic setup of these concepts. Thanks for the help in advance! 
 A: Injective: Choose any $x_1, y_1, x_2, y_2 \in \mathbb Z$ such that $f(x_1, y_1) = f(x_2, y_2)$ so that:
\begin{align*}
5x_1 - y_1 &= 5x_2 - y_2 \\
x_1 + y_1 &= x_2 + y_2
\end{align*}
We want to show that $x_1 = x_2$ and $y_1 = y_2$. Hint: To prove the first part, begin by adding the two equations together. The second part follows by substitution.

Surjective: Choose any $a,b \in \mathbb Z$. We seek some $x, y \in \mathbb Z$ such that $f(x, y) = (a, b)$ so that:
\begin{align*}
5x - y &= a \\
x + y &= b
\end{align*}
Hint: As before, add the two equations together to solve for $x$ in terms of $a$ and $b$. Then substitute to solve for $y$ in terms of $a$ and $b$.
A: Note that this function isn't surjective.  Note that if we write $f(x,y) = (a,b)$, then $a + b = (5x-y) + (x+y) = 6x$.  This means that the image does not contain any $(a,b)$ except those for which $a+b$ is divisible by $6$.  In particular, $f(x,y) \ne (1,0)$ for any $x,y\in \Bbb Z$.
To make a general statment, the determinant of the matrix $\begin{bmatrix}5 & 1 \\ -1 & 1 \end{bmatrix}$ is $6$ instead of $\pm 1$, so the corresponding linear map $\Bbb Z^2 \to \Bbb Z^2$ will not be surjective.
This can be fixed if you replace the $\Bbb Z$'s with $\Bbb Q$'s in the domain of $f$.  
