Fixed Point Theorem and Closed Unit Ball in Euclidean Space

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The Brouwer Fixed-Point Theorem

Let denote the closed unit ball in Euclidean space :
.
Any continuous map from onto itself has at least one fixed point, i.e. a point such that .

Proof Suppose has no fixed points, i.e. for .
Define a map , , by letting be the point of intersection of and the ray starting at the point and going through . For see figure below:

We have
with , (1)
Then
(2)
and so is continuous. Could you please explain, in as much detail as possible, how (1) and (2) were derived?

EXPLANATION: Since be the point of intersection of and the ray starting at the point and going through .
So r(x) is a point on the surface of Hollow sphere. Which is the point of intersection.

Since it is a point of intersection . Therefore r(x) is a point. Also given that r(x ) lies on
Sn-1(x) . And it is also specific that

so we can easily imagine that

Any point on Sn-1 (x) is in such a way that x belongs to R n such that mod (x )= 1.

So clearly any point on Sn-1 (x) is in such a way that

with ,

Also it is great that r(x) is a function in such a way it generates the points which are on the Hollow ball(sphere)(because mod (x) = 1) with the help of m and t.

Where these m and t are just defined functions or expressions or standard expressions

and

Here m is taken in order that r(x) will lie on Sn-1(x).

Also t is taken in such a way that totally r(x) will lie on Sn-1(x).

So here Sn-1(x) is a hollow ball without any material in it.It just contains the over surface to cover like a ball.

These are the explanations according to my point of view.

CLARIFICATION PART:

As with ,

Also x + mt = 1 implies x + mt = +1 or - 1.

Implies x + mt = 1 or x + mt = -1

Implies mt = 1 - x or t = ( 1 - x )/m

ANOTHER Equation ...