Homology group in Hatcher I rather not write out the entire essay discussion on Hatcher on page 100, Chapter 2 on Homology. But we have a graph that looks like the following.



*

*In this discussion $a-b$ is the boundary of $A$ and Hatcher argues that by "sliding" the boundary into $A$, $a-b\simeq \{e\}$, basically the 2-cell is convex is what is he saying. So now he argues that we quotient out the cycle $a-b.$ And Hatcher say that in this quotient, $a-c \simeq b - c.$ I am not sure what this precisely means here. What are the other equivalences in this graph?

*Next he officially introduces the (first) homology group $H_1(X_2) = Ker(\partial_1 :C_1 \to C_0)/Im(\partial_2: C_2 \to C_1) = Ker(\partial_1)/\{e\} = \{ n(b - c)+ m(c - d) + 0\} \simeq \mathbb{Z} \times \mathbb{Z}$ because $\{ e \} = \{ a - b \}$. This part isn't a question, just "summarizing" in case you don't want to pull up his book.

*Now we move our case up to this picture. My confusion starts here. For one, I don't quite understand what I am looking at here because Hatcher says $\partial_2$ sends both $A$ and $B$ to $a-b$. Now it looks like to me the boundary isn't actually just the cycle $a-b$, what about those "boundary circles" he later mentions? Why aren't those the boundary of this?

*Next he claims $H_1(X_3) = H_1(X_2) = $, assuming I agree with his previous statement, this is true. Now he claims $Ker(\partial_2)$ is nontrivial and is generated by $A-B$. Was $Ker(\partial_2)$ trivial before? Because $Ker(\partial_2) = A$ when we were in $X_2$ since we agreed that $a-b \simeq \{e\}$. Also what is meant by $A-B$? Seems like the subtraction induces some orientation this new cell.

*Next we conclude that $H_2(X_3) = Ker(\partial_2)$ This is because we are taking $C_3 = \{0\}$. But it isn't quite clear to me that $Ker(\partial_2) \simeq \mathbb{Z}$ since $\partial_2(nA + mB) = (a-b) \implies Ker(\partial_2) = \{nA + mB \} \simeq \mathbb{Z} \times \mathbb{Z}$ or am I missing something here? I am going to stop here because this post already long enough and I didn't go any further with this.

 A: 
1 . "Hatcher say that in this quotient, $a-c \simeq b - c.$ I am not sure what this precisely means here."

What Hatcher is saying is that following along $a$ and then reverse along $c$ is equivalent to following along $b$ and then reverse along $c.$ This is evident since $a-b \simeq e,$ that is $ a-b\in \text{Im}(\partial_1),$
so there exists some $z\in  \text{Im}(\partial_1)$ so that $a-b=z,$ or more conveniently $b=a-z.$ The point being is that if  $[a]$ and $[b]$ are the representatives associated to $a$ and $b,$ modulo $\text{Im}(\partial_1),$ then $[a]=[b].$ So we could write this a little less carefully as $a-c=b-c$ modulo $\text{Im}(\partial_1).$

What are the other equivalences in this graph?

Run the same argument using $d$ instead of $c.$


  
*($X_2$ the space with two two-cells $A,$ and $B$ attached.) Hatcher says $\partial_2$ sends both $A$ and $B$ to $a-b$. Now it looks like to me the boundary isn't actually just the cycle $a-b$, what about those "boundary circles" he later mentions? Why aren't those the boundary of this?
  

Before you think of both $A$ and $B$ being attached, think of attaching just $A.$ What is the boundary? The graph indicates $a-b.$ Do the same for $B.$ Now that both have been glued on, you'll see that $\partial_2$ sends $A$ and $B$ to $a-b.$ You're right, in a sense, that the boundary "isn't just" $a-b.$ However, once you've specified the $1-$cell structure, you only need to establish the boundaries you're gluing your $2-$cells to. In this case, we're gluing along $a-b.$ If you view the space, without specifying a $1-$cell first, then you could argue that some other loop on the surface of the (homeomorphic) copy of $S^2$ is the boundary, but as long as you specify $a-b$ first, then that's the boundary you'd like to think about. Any other boundary is, of course, homotopic to $a-b.$ 


  
*Next he claims $H_1(X_3) = H_1(X_2)$, assuming I agree with his previous statement, this is true. Now he claims $Ker(\partial_2)$ is nontrivial and is generated by $A-B$. Was $Ker(\partial_2)$ trivial before?
  

Yes, it was trivial before, since $\partial_2(A)=a-b\ne0.$ 

Because $Ker(\partial_2) = A$ when we were in $X_2$ since we agreed that $a-b \simeq \{e\}$. 

The statement immediately previous to this one doesn't make sense, since $\partial_2(A)=a-b\ne0.$ The loop $a-b$ is in the kernel of $\partial_1.$

Also what is meant by $A-B$? Seems like the subtraction induces some orientation this new cell.

It does, to an extent. Call any circle passing through $x$ and $y$ as in the graph, "vertical." By the fact that any of these loops must be homotopic to $a-b,$ they have the same orientation. 
On the other hand, take any path from some fixed point on the path $a$ to a fixed point on the path $b,$ you may think of these as "horizontal" circles." Let's call this path $p_A,$ to denote that it lies on the $2-$cell $A.$ We'll then assume that any such path, from $a$ to $b,$ is positive. Similarly, define a new path $p_B,$ which lies on the $2-$cell $B,$ and has the same positive orientation, namely, traversing from $a$ to $b.$ Then there is a loop $p_A-p_B.$ Giving an orientation one might associate to the difference $A-B.$ 


  
*Next we conclude that $H_2(X_3) = Ker(\partial_2)$ This is because we are taking $C_3 = \{0\}$. But it isn't quite clear to me that $Ker(\partial_2) \simeq \mathbb{Z}$ since $\partial_2(nA + mB) = (a-b) \implies Ker(\partial_2) = \{nA + mB \} \simeq \mathbb{Z} \times \mathbb{Z}$ or am I missing something here? 
  

The boundary map $\partial_2(A)=a-b,$ and $\partial_2(B)=a-b,$ gives us $\partial_2(A-B)=0.$ So $\ker(\partial_2)=\mathbb{Z}\{A-B\},$ since $$\partial_2(nA-mB)=n\partial_2(A)-m\partial_2(B)=n(a-b)+m(a-b)\ne (a-b)$$ (in general).
You can see that $\partial_2(A)\ne 0,$ and $\partial_2(B)\ne 0,$ but $\partial_2(A-B)=0,$ so $A-B\in \ker(\partial_2).$ Since $C_2 = \mathbb Z\oplus \mathbb Z,$ $\partial_2(A+B) \ne 0,$ and $\{A+B,A-B\}$ form a $\mathbb Z-$basis, then $\ker(\partial_2)$ and $\text{Im}$ are both isomorphic to $\mathbb{Z}.$
