In Brouwer's theorem, how does the absence of fixed points imply certain retraction properties? I'm trying to understand the proof of Brouwer's theorem form Lawvere and Schanuel's Conceptual Mathematics:

To prove that the non-existence of a retraction implies that every
  continuous endomap has a fixed point, all we need to do is to assume
  that there is a continuous endomap of the disk which does not have any
  fixed point, and to build from it a continuous retraction for the
  inclusion of the circle into the disk.

This seems to involve unspoken assumptions that I don't know how to justify. If the non-existence of the retraction proves $ \exists x[f(x) = x] $, then I believe that it must also be true that the lack of fixed points $ \forall x[f(x) \neq x] $ implies the existence of a retraction which shares the same property as the endomap, namely:
$$∀x[(f(x) \neq x)] \to \exists r[r \circ j = 1_C]$$
(Edit note: this expression was originally $∀x[(f(x) \neq x) \to (r(x) \neq x)]$. I mistakenly thought that contradicting $ r(x) \neq x$ was necessary to the argument.)
So my question is twofold:


*

*How does the lack of a fixed point in the endomap imply (as opposed to simply making possible) the existence of a retraction, $ r \circ j = 1_C $?

*Why is it assumed that the retraction must have properties that behave as the endomap lead to a contradiction? In other words, it seems like they could be completely independent so that it's both true that $ \forall x[f(x) \neq x] $ and $ r \circ j = 1_C $.


Here's the proof from the book:


So, let $ j: C \to D $ be the inclusion map of the circle into the
  disk as its boundary, and let's assume that we have an endomap of the
  disk, $ f: D \to D $, which does not have any fixed point. This
  means that for every point $x$ in the disk $D$, $f(x) \neq x.$
From this we are going to build a retraction for $j$, i.e. a map $r: D
\> \to C$ such that $r \circ j$ is the identity on the circle. The key to
  the construction is the assumed property of $f$, namely that for every
  point $x$ in the disk, $f(x)$ is different from $x$. Draw an arrow
  with its tail at $f(x)$ and its head at $x$. This arrow will 'point
  to' some point $r(x)$ on the boundary. When $x$ was already a point on
  the boundary, $r(x)$ is $x$ itself, so that $r$ is a retraction for
  $j$, i.e. $rj= 1_c$.

(The book doesn't assume any knowledge about topology nor do I know much about it.)
 A: You can't construct a retraction if there is a fixed point because the basic ingredient in the construction of the retraction is the displacement direction $f(x)-x$, which can be normalized to get a unit vector. If the displacement is zero then you can't get a unit vector because you can't divide by zero. In more detail, with regard to your question

How does the lack of a fixed point in the endomap imply the existence of a retraction, $ r \circ j = 1_C $?

The answer is that the absence of a fixed point of the endomap is what allows one to define a continuous retraction.  With regard to your question 

Why is it assumed that the retraction must behave as the endomap? In other words, it seems like they could be completely independent so that it's both true that $ \forall x[f(x) \neq x] $ and $ \exists x [r(x) = x]$

The retraction itself (when it is defined, i.e. so long as the endomap has no fixed points) satisfies $r(x)=x$ for all points $x$ on the boundary, by construction of $r$.
As I mentioned in the comments, the account provided by the authors you cite leaves an essential topological fact unmentioned. This is the fact that the identity map of the circle cannot be extended over the disk.
A: The claim made is essentially: suppose we can show that there is no continuous retraction from $B^n$ onto $S^{n-1}$ (you would use homotopy or homology to show that, usually).
Then every continuous $f: B^n \to B^n$ (an endomap) has a fixed point.
Reason: if we have any such endomap $f$ without a fixed point, then define $r(x)$ as the unique point of intersection of the ray $x + t(f(x) -x), t \in [0,\infty)$ with $S^{n-1}$ having the smallest $t$, is well-defined (as $f(x) \neq x$ on $B^n$, we always have a non-zero vector as a direction for the ray), continuous (hint here, e.g.) and a retraction from $B^n \to S^{n-1}$, (for $x \in S^{n-1}$ $t=0$ is the first intersection point, so $r(x) = x$) which we have shown to be impossible.
Note that also that having no such retraction is a necessary condition for the fixed point property: if we'd have such an $r$ we could define $f = h\circ r: B^n \to B^n$, where $h$ is a non-trivial rotation of $S^{n-1}$,and this $f$ has no fixed point (points inside map to points on the boundary, so are never fixed, and on the boundary $r$ keeps them fixed but $h$ moves them). 
So the fixed point property for $B^n$ (i.e. Brouwer's theorem) is equivalent to the no-retraction theorem, in the sense that one is easily derived from the other.
A: The claim being made is actually (roughly)
$$(\exists r)(r\,\mathrm{is\,a\,retraction})\implies (\forall f)(\exists x)(f(x)=x)$$
which is distinct from the formula you gave. I do not know if this is the source of your confusion, but it's worth pointing out. 
For your first question, suppose $f:B^n\to B^n$ has no fixed point. When we may define $r: B^n\to S^{n-1}$ as follows: given $x\in B^n$, consider the unique line intersecting both $x$ and $f(x)$ (this line is unique because $f(x)\neq x$). Then define $r(x)$ to be the point of intersection of this line with the sphere, on the side of $f(x)$. This defines a continuous retraction from the ball to its boundary. 
Thus, proving that no retraction exists from the ball to its boundary is sufficient to prove the Brouwer fixed point theorem. For if there were a map with no fixed point, as shown, we could contruct such a retraction. This should clarify your second question as well. 
