cokernel of a homomorphism I am dealing with the following problem:
Given $f:R \rightarrow R^2$ such that $f(x)=(x,-x)$ I am trying to compute the quotient group of $f$.
So I know that this is the group $R^2/Im(f)$.
I find that $Im(f)=\{(x,y)\in R^2 / y=-x\}$
Then, I dont know how to continue...
If i am applying the definition correctly, it should be the set $\{z+x,u-x\}$ for $(z,u) \in R^2$  but I dont think that this is the proper result. Any advice for this?
Moreover, I am attending lectures for Algebraic Topology and I realize that I am weak in Algebra (I did the very basics of group theory as an undergaduate student). Would you suggest me to study simultaneously group theory in order to keep up with the Alg. Topology?
 A: We can assume $R$ is any Abelian group here.
Consider the homomorphism $g:R^2\to R,\ g(x,y)=x+y$.
Its kernel is just ${\rm im}(f)$ and it's surjective, as $g(x,0)=x$ for any $x\in R$, hence by the first isomorphism theorem,
$$R^2/{\rm im}(f)\,\cong\,R\,.$$
Alternatively, one can directly show that $R$ with map $g$ satisfies the universal property of cokernel: $g\circ f=0$ and  any group homomorphism $h:R^2\to S$ with $h\circ f=0$ uniquely factors through $g$.
A: Thanks for your answers so far.
In order to not spam with new posts, I would like to give my solution to a similar problem and I would like your feedback about it if that is possible.
I want to find the isomorphism classes of the quotient group of the homomorphism
$f:\mathbb{Z} \bigoplus \mathbb{Z}_{2} \rightarrow \mathbb{Z} \bigoplus \mathbb{Z}_{4}$, with $f(n,m)=(2n,2m)$.
I compute that $Im(f)=\{(x,y)\in \mathbb{Z} \bigoplus \mathbb{Z}_{4}/x=2k, k \in \mathbb{Z} , y \in \{\bar{0}_{4},\bar{2}_{4}\}\}$.
So in order to compute the $\mathbb{Z} \bigoplus \mathbb{Z}_{4}/Im(f)$ I have to compute the
$aIm(f)=\{a+h:h\in Im(f)\}$ for every $a\in \mathbb{Z} \bigoplus \mathbb{Z}_{4}$.
Now $a$ can be $(z,\bar{0}_{4}),(z,\bar{1}_{4}),(z,\bar{2}_{4}),(z,\bar{3}_{4})$ where $z\in \mathbb{Z}$
Doing the computation, I find that the quotient group is $\{(\mathbb{Z},\{\bar{0}_{4},\bar{2}_{4}\}),(\mathbb{Z},\{\bar{1}_{4},\bar{3}_{4}\})\}$
But considering the set $\{\bar{0}_{4},\bar{2}_{4}\}$ I observe that this is the $\bar{0}_{2}$ and that $\{\bar{1}_{4},\bar{3}_{4}\}$ is the $\bar{0}_{4}$. Now noticing that every element of $\bar{0}_{4}$ is into $\bar{0}_{2}$, I find that the group  $\mathbb{Z} \bigoplus \mathbb{Z}_{4}/Im(f)$ is $\mathbb{Z} \bigoplus \{\bar{0}_{2}\}$
Is the solution correct?
One remark. Considering that the cokernel of a function shows the "divergence" of a map of being surjective, I could consider that the result is correct because: $f$ "cuts" the odd integers (thats why $\mathbb{Z}$) and also "cuts" the elements $(2,6,10,14,...)$ and these elements belong in $\bar{0}_{2}$. Does this way of thinking make sense?
Thanks in advance.
A: I want to find the isomorphism classes of the cokernel of $f$. I think i divided by the image, but maybe I did something wrong during the further computations.
EDIT: Actually I think that I have found my mistake. So, I have that $Im(f)=\{(x,y)\in \mathbb{Z} \bigoplus \mathbb{Z}_{4}/x=2k, k \in \mathbb{Z} , y \in \{\bar{0}_{4},\bar{2}_{4}\}\}$ and when I divide $\mathbb{Z} \bigoplus \mathbb{Z}_{4}$ by $Im(f)$ I obtain for the first part $\mathbb{Z}/2k$ (which is $\mathbb{Z}_{2}$) and for the second part the $\{\bar{0}_{4},\bar{2}_{4}\}$ (which is $\bar{0}_{2}$) and $\{\bar{1}_{4},\bar{3}_{4}\}$ (which is $\bar{1}_{2}$) so the second part is again $\mathbb{Z}_{2}$. So the cokernel of $f$ is indeed $\mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2}$.
I hope that now it is ok. Thank you very much for your help, things start getting clear about that now.
