proof a function is surjective

Question

Most of the Time in order to proof a function is surjective

• we first choose an Arbitrary point belong to Co-domain (Target Point)
• Let (Target Point) = definition of function
• Solve for x
• proof that f(x) = y

in example Like :

f : Z×Z→Z×Z defined by the formula f(m,n)= (m+n,m+2n)

My Proof was as the Following :

Let (x,y) ∈ Z×Z #(this is the target point)

x = m+n solve for n thus n= x-m

y = m+2n solve for m thus m=y-2n

f(m,n) = (x,y)

(m+n,m+2n) = (x,y)

(y-2n + n , y-2n+2n)

(y-n , y) = (x,y)

I'm stuck here I don't Know I should do to get y-n = x

Answer 1

The function is just a multiplication by a matrix $\begin{pmatrix} 1 & 1\\1& 2 \end{pmatrix}$, which has the inverse $\begin{pmatrix} 2& -1\\-1& 1 \end{pmatrix}$ (over the reals, but it works over $\mathbb{Z}$ too, as we have integer coefficients and the determinant $1$).

So to solve $f(x,y) = (m,n)$ we take $x=2m -n, y=-m+n$, and you can just compute the images to see that it checks out that $f(x,y) = (m,n)$.

Answer 2

Subtract the first equation from the second.
Thus y - x = n.
Use that value for n in the first equation and solve for m.
Have you not learned how to solve simultaneous equations?

Answer 3

2 linear equations, 2 unknowns. So solve the system of equations. (Just like you did in algebra I).

$m + n = x$.

$m + 2n = y$

x = m+n solve for n thus n= x-m

y = m+2n solve for m thus m=y-2n

And thus $m = y- 2(x - m)$

.....

So $m = 2x - y$

And $n = x-m = x - (2x - y) = y-x$.

So $f(2x - y, y-x) = ((2x -y) +(y-x), (2x-y) + 2(y-x)) = (x, y)$.

That's it.