I'm currently stuck on a math problem, here is the instruction :
"Let R be the equivalence relation on N × N defined as follows: (m1, n1) ∼ (m2, n2) if m1 + n2 = n1 + m2.
Let f : N × N → N × N/R denote the quotient map, and consider the function g : N × N → Z defined as g(a, b) = a − b. Show that g can be identified with the quotient by R: that is, show that there exists a bijection h: N × N/R → Z"
I have proven the existence of h using the universal property of the quotient (because h ◦ f = g). But I can't quite prove that it is a bijection. I have tried to prove the injectivity of h by proving the injectivity of g, but somehow I can't prove its injectivity (maybe g isn't injective ?). And for the surjectivity I have no idea. Or maybe I should prove the bijection using cardinals ?
I'm a 1st year maths student btw, thank you in advance :)
Summarizing what's been covered in the comments:
First for $a,b,c,d\in\mathbb{N}$ since $(a,b)\sim(c,d)$ means $a+d=b+c$, which implies $$g(a,b)=a-b=c-d=g(c,d)$$ it follows from the universal property of the quotient map $f:\mathbb{N}^2\to\mathbb{N}^2/{\sim}$ that there is a unique map $h:\mathbb{N}^2/{\sim}\to\mathbb{Z}$ with $g=h\circ f$.
To prove that $h$ is surjectve it is sufficient to prove that $g$ is surjective, because if $y=g(a,b)$ then $y=h(f(a,b))$. But $g$ is surjective, since for any $z\in\mathbb{Z}$ we can find $a,b\in\mathbb{N}$ with $a-b=z$ and hence $g(a,b)=z$. For example if $z\ge0$ we can take $a=z$ and $b=0$, and if $z<0$ we can take $a=0$ and $b=-z$. In fact there are infinitely many possible choices of $a$ and $b$.
To prove that $h$ is injective, suppose that $h(f(a,b))=h(f(c,d))$. Then $$a-b=g(a,b)=g(c,d)=c-d$$ so $a+d=b+c$, which means $(a,b)\sim(c,d)$, which implies $f(a,b)=f(c,d)$ by definition of the quotient map $f$. The injectivity of $h$ now follows since $f$ is surjective.