I'm trying to compute the cokernel of $\mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z}$ where $(a,b) \rightarrow (a+b, a-b)$. I realise that the image of this map is just $\mathbb{Z} \oplus \mathbb{Z} - \{(a,b) \in \mathbb{Z} \oplus \mathbb{Z} : a \not\equiv b \text{ mod } 2\}$
Am I right in thinking that its cokernel will just be $\{(0,0), (1,0), (0,1) \}$, I'm not exactly sure how to write this, but my thinking is that the cokernel will be the quotient of $\mathbb{Z} \oplus \mathbb{Z}$ by the image. This will mean that the elements of the cokernel are not in the image and the different equivalence classes can't be related by an element in $\mathbb{Z} \oplus \mathbb{Z}$ with $a + b \equiv 0 \text{ mod }2$
Is this correct and if so, is there some better notation I can use to represent this cokernel?
The image, as you say, is the set $H=\{(a,b):a\equiv b\pmod 2\}= \{(a,b):a+b\text{ is even}\}$. The cokernel is $(\Bbb Z\oplus\Bbb Z)/H$ which consists of the cosets of $H$ in $\Bbb Z\oplus\Bbb Z$. There are two of these: $H$ and $(1,0)+H=\{(a,b):a+b\text{ is odd}\}$. So the cokernel is the group of two elements.