Deformation retraction arising from a mapping cylinder. In Allen Hatcher Algebraic topology, chapter zero, I had the following:





My questions are:
1- Why the mapping cylinder is the quotient space, what makes it specifically a quotient space?
2- Why there is the part I circled in red in the picture in the mapping cylinder, what is its importance?

 

 A: A quotient space is what happens when you take a topological space, and identify some of its points. The mapping cylinder is defined as the given disjoint union, with some points identified. That makes it a quotient space.
The author is assuming the reader knows the term from earlier topology studies, and uses it in order to avoid having to write out the formal construction in all its details. (Hatcher is known for including fewer details and less rigor in this book than many would have preferred, but the exercises are virtually unsurpassed.)
The bendy bit on the drawing is there to give a visual clue as to what this quotient looks like. They could've made the drawing without it, but the result would've been less interesting, and frankly, less enlightening.
A: I'm not sure how to interpret the first question, but there are a few answers.
1) It is a quotient space because it is a quotient. You define the mapping cylinder of $f\colon X\to Y$ as the quotient of the space $X\times[0,1]\amalg Y$ by the relation $(x,1)\sim f(x)$.
2) A surjective map $q\colon A\to B$ is a quotient map if and only if "a set $U$ is open in $B$ if and only if $q^{-1}(U)$ is open in $A$." In other words, $B$ has the quotient topology. In this case, the quotient map is $X\times[0,1]\amalg Y\to(X\times[0,1]\amalg Y)/\sim$.
3) In another sense, to "be a quotient space" means nothing. Any space $X$ is a quotient space of itself with the trivial relation.
As to the second question, there is no significance to the circled area. Hatcher just tried to draw something more interesting than a circle.

The point of mapping cylinders is really so that, for any map $f\colon X\to Y$, you can replace $f$ with the inclusion of $X$ into the mapping cylinder $M_f$. Here, "inclusion" means the map $i\colon X\to M_f$ sending $x$ to $(x,0)\in X\times[0,1]$. Then via a straight-line deformation, $M_f\simeq Y$. The point is that, when you want to compute the homology of one of the spaces $X$ or $Y$, you can now think of $X$ (i.e. $X\times\{0\}$) as being a subspace of $Y$ (i.e. $M_f$), and then look at the long exact sequence of the pair $(M_f,X)$.
