Questions about quotient space and homology I watched the lecture 31 of algebraic topology on youtube and could not figure out a problem. The link and the question picture are as follows: 
link:https://www.youtube.com/watch?v=2wn10l9qbJI at time 21:32 

 
(In this picture the professor should wrote $c-d$ instead of $c+d$) 
The professor tried to say how to understand the homology means when a 2-cell is added to the 1-dim graph using a example as the picture shows. Before adding the 1-dim graph, I can understand the homology of the graph is just Z $\times$ Z because in this graph we can find two basic cycles ( or called generators). Then for every other cycles in this graph, we can represent it by the linear combinations of the two basic cycles. But when the professor added the 2-cell, he told that the homology became Z$\times$Z/Z which is isormophic to Z. Then he tried to tell that what is a quotient means but I can not understand. I think I can summary my question as follows: 
1. Is the answer of quotient operation X/A the set of cosets of A? For example if $Z$ represents the integer, so $Z$/$3Z$ just equals to 0,1,2 the 3 elements?
2. If what I thought about question 1 is right, in this case, why <$a+b+c$, $a+b+d$>/<$c-d$>, which can be written as $Z$ $\times$ $Z$/$Z$, is $Z$? I mean it can be many cosets here. For example, <$2a+2b+c+d$>, <$3a+2b+2c+d$>, <$a+b+2c-d$>. And maybe why $a+b+c$ and $a+b+d$ are in the same coset is that ($a+b+c$) - ($a+b+d$) equals to $c-d$ so they are in the same coset. But we want is <$a+b+c$, $a+b+d$>/<$c-d$> and I think the answer should be multiple cosets, like 0,1,2 in the quetion 1. More precisely, I think I can view <($a+b+c$) - ($a+b+d$)> as points <(1,-1)> in $Z^2$, such as (2,-2), (3,-3). But I can still get subgroups such as <(2,3)>,<(1,5)> and so on. So I think quotient answer should not be $Z$. It should also be $Z \times Z$ 
So could anyone help me out?
 A: 
  
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*Is the answer of quotient operation X/A the set of cosets of A? For example if $Z$ represents the integer, so $Z$/$3Z$ just equals to 0,1,2 the 3 elements?
  

If you have a set $X$ and an equivalence relation $\sim$ then you have
$$[x]_{\sim}=\{y\in X\ |\ y\sim x\}\text{ for }x\in X$$
$$X/\sim=\big\{[x]_{\sim}\ |\ x\in X\big\}$$
This $X/\sim$ is the quotient set.
Now if $A\subseteq X$ then the quotient space $X/A$ has to be defined as above. In other words you have generate a relation $\sim$ from $X$ and $A$.
Things get more complicated now because the definition of $\sim$ depends on the context, for example $X/A$ does not mean the same thing in the group theory and in the topology.
In topology you usually have that $X/A=X/\sim$ where $\sim$ is defined as follows: $x\sim y$ if and only if $x=y$ or $x,y\in A$. In other words $X/A$ is the result of a process of collapsing a subspace.
In algebra if $G$ is a group/ring/module and $H\subseteq G$ is a subgroup/ideal/submodule then $G/H$ means something different. Now $G/H=G/\sim$ where $x\sim y$ if and only if $xy^{-1}\in H$ (or in abelian case $x-y\in H$). This definition coincides with the definition by cosets. In terms of cosets you would just write $[x]_{\sim}=x+H$. And here your intuition is almost correct. If $\mathbb{Z}$ is the group of integers, then
$$\mathbb{Z}/3\mathbb{Z}=\big\{[0], [1], [2]\big\}$$
$$\text{or}$$
$$\mathbb{Z}/3\mathbb{Z}=\big\{0+3\mathbb{Z}, 1+3\mathbb{Z}, 2+3\mathbb{Z}\big\}$$
$$\text{or even simplier (if doesn't lead to confusion)}$$
$$\mathbb{Z}/3\mathbb{Z}=\{0,1,2\}$$
is the group of integers modulo $3$. Note that the last equality is not really equality. Is just a shortcut. First two are proper definitions.

I mean it can be many cosets here. For example, $<2a+2b+c+d>$, $<3a+2b+2c+d>$, $<a+b+2c−d>$.

These are not cosets. These are subgroups. To form a coset you need a subgroup $H\subseteq G$. Then a coset is a subset of $G$ of the form $gH$ (or in abelian case $g+H$). These are almost never subgroups! In fact these are subgroups if and only if $g\in H$.


  
*If what I thought about question 1 is right, in this case, why <$a+b+c$, $a+b+d$>/<$c-d$>, which can be written as $Z$ $\times$ $Z$/$Z$, is $Z$?
  

Forget about $a,b,c$ and $d$. These are just labels and addition as above is not really an addition. By $a+b$ he actually understands an abstract object in some abstract group. All of that is just a distraction.
It would be easier for you to think about it as follows: we have one generator $A=a+b+c$ and second generator $B=a+b+d$. Together they form a group $<A,B>$ which happens to be $\mathbb{Z}\oplus\mathbb{Z}$. In that group we have a subgroup generated by $A-B=c-d$ (yes, there's a typo on the board, you cannot generate $c+d$ from $a+b+c$ and $a+b+d$, hence it makes no sense to talk about quotient). 
Now this is still too abstract. Let's rewrite this even further:
$$G=\mathbb{Z}\oplus\mathbb{Z}$$
$$A=(1,0)$$
$$B=(0,1)$$
$$A-B=(1,-1)$$
$$H=\langle(1,-1)\rangle$$
What's the quotient group $G/H$? Consider the function
$$f:\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}$$
$$f(x,y)=x+y$$
You can easily check that this is a group homomorphism that is surjective. Note that the kernel of $f$ is equal to $H$. It follows (by the First Isomorphism Theorem) that $G/H\simeq\mathbb{Z}$.
In other words every coset in $G/H$ can be uniquely written in the form $(n,0) + H$ for $n\in\mathbb{Z}$.
