Questions for two exercises on visualising quotient spaces I am attempting to do the following two visual exercises in quotient spaces.  With the exception of the last one for the second question, I have listed what I think their respective equivalence classes are.  For some of them, I can only guess what the quotient space look like. I am having trouble picturing in the plane what it means visually for two separate to be consider identical.  I am hoping someone from the community can provide me with some much needed clarifications. Thank you in advance.
1) Suppose $X=\mathbb{R}^2$.  Describe (visualize) the space $X/\text{~}$ if ~ is the smallest equivalence relation satisfying the following conditions. 
a) $(x,y)\text{~}(x',y')$ if $x=x'-1$ 
The equivalence class is $[(x'-k,y')]=\{(x',y'),(x'-1,y'),(x'-2,y'),...(x'-k,y')....\}$.  I am guessing this is a cylinder 
b) $(x,y)\text{~}(x',y')$ if $x=x'-1$ and $y=y'-1 $
The equivalence class is $[(x'-k,y'-k)]=\{(x',y'),(x'-1,y'-1),(x'-2,y'-2),...(x'-k,y'-k)....\}$ I am guessing this is a torus.  Because 
c) $(x,y)\text{~}(x',y')$ if $x=x'-1$ or $y=y'-1 $
The equivalence class are: 
$[(x',y'-k)]=\{(x',y'),(x',y'-1),(x',y'-2),...(x',y'-k)....\}$ or
$[(x'-k,y')]=\{(x',y'),(x'-1,y'),(x'-2,y'),...(x'-k,y')....\}$ or 
$[(x'-k,y'-k)]=\{(x',y'),(x'-1,y'-1),(x'-2,y'-2),...(x'-k,y'-k)....\}$

2) Denote the  usual, $S^2=\{(x,y,z)\in\mathbb{R}^3:x^2 + y^2 + z^2 =1\}$ (the unit sphere), and $D^3=\{(x,y,z)\in\mathbb{R}^3:x^2 + y^2 + z^2 \leq1\}$ (the closed unit ball).
a) Define $f:S^2\rightarrow S^2$ by $f(x,y,z)=(x,y,-|z|)$.   Describe the space $S^2/\text{~}_f$
Is the equivalence relation the following:  (x,y,z)~(x,y,-z)?  I don't know what the quotient space suppose to look like.
b) Define $g:D^3\rightarrow D^3$ to be $f$ over $S^2$ and the identity elsewhere..   Describe $D^3/\text{~}_g$
I am not sure how g is defined. I don't understand what it meant by "$f$ over $S^2$ and the identity"
 A: 1) a) Indeed a cylinder, which loops at every $[(0.9999, y)]$ to $[(0., y)] = [(1., y)]$. This divides $R^2$ into long vertical horizontally looping strips. By identifying them, (ie, saying they are all the same and keeping just one), you indeed get something homeomorphic to a cylinder (infinite in a direction).
You could also say your equivalence is generated by $(x', y) - (1, 0) \equiv (x, y)$
1)b) Here you divide your space both vertically and horizontally, that gives you a tiling of squares that loop on each side. Indeed they are all homeomorphic to torii (look up the concept of "fundamental polygon", the one for the torus is exactly your case).
2) For a quotient of a space by a function, you have to think of it as "two element $a$ and $b$ are equivalent/congruent (ie, $a \equiv b$) iff $a - b \equiv 0$". So taking  $f(x, y, z) = (x, y, -|z|)$, you can see:
$f(u) \equiv f(u')$ iff 
$f(x, y, z) - f(x', y', z') \equiv (0, 0, 0)$ iff
$(x - x', y - y', -|z| - (-|z'|)) \equiv (0, 0, 0)$ iff
$x = x'$ and $y = y'$ and $|z| = |z'|$
So you're basically identifying the points of the sphere through a mirror symmetry with the points on your mirror/$xy$-plane as your invariants (ie, the equator). This gives your something like a disc, but with particular edges. Within this disc, particles would reflect at the edges, like pool balls would bounce at a circular pool table. I don't know if it's homeomorphic to some better known topological structure. It's not $RP^2$ because $RP^2$ seen as a disc would warp the particles to the diametrically opposed edge of the disc.
2)b) I suppose you mean "$S^2$ and the identity". Basically, the outer boundary points of your ball are "linked" ("identified", made to be identical) as they are above in 2)a), and there is no new linking happening in the points interior to the ball (the quotient of any topological space by the identity function is the space itself). The trick here is to realize that paths starting on the "upper hemisphere" can now attain points on the lower hemisphere not only through the equator, but also by going through the interior of the earth. This is a bit hard to think about. My gut tells me it's something like $RP^3$, but I'm not at all convinced and wouldn't be sure how to prove that. Tell me if you find a better answer to this one. 
As a final note, a good way of understanding quotients is basically "injecting a desired property" (an equality defined by an equivalence relation) into a space. Though you have to verify that the equivalence class gives birth to a "well-defined quotient". You have to ensure some category-theoretical commutativity relations. For example, the working quotients of vector spaces have to maintain coherence of restriction to equivalence classes, as well as linearity between an initial space V and its quotient W: 
$(\forall x, y \in V, x \equiv y \iff 
[x] = [y] $ and [x], [y] are well defined as elements of a vector space$)$
$\iff$
$(\forall \lambda \in K, x, y \in V, f_W([x + \lambda y]) = [f_V(x)] + \lambda [f_V(y)])$
where $f_W : W \to A$ is the obvious restriction of $f_V: V \to A$ which takes $V$ in input to the quotient space $W$ (composed of identified elements of V) as input, for any output vector space $A$.
