I will appreciate if someone could explain to me the solution of the following problem:
The unit ball in $\mathbb{R}^{n}$ and a point are homotopically equivalent.
Def 1: Two spaces $X$, $Y$ are homotopically equivalent if there exists $f:X\to Y$ and $g:Y\to X$ so that $f\circ g$ is homotopic to $id_{y}$ and $g\circ f$ is homotopic to $id_{x}$
Def 2: A homotopy between two continuous maps $f,g:\;X\to Y$ is a continuous function $H\;:X\times [0,1]\to Y$ such that if $x\in X$ then $H(x,0)=f(x)$ and $H(x,1)=g(x)$
Solution:
Let $X$ be the unit ball and let $Y:=\{\bullet\}$. Let $f:X\to Y$ be defined as $f(x)=\{\bullet\}\;,\forall x\in X$ and let $g:Y\to X$ so that $g(\{\bullet\})=0$.
Then $\begin{cases}f\circ g:Y\to Y\\(f\circ g)(\{\bullet\})=\{\bullet\}\end{cases}\;\;$ so $f\circ g=id_{Y}\;$ and $\begin{cases}g\circ f:X\to X\\(g\circ f)(x)=0\end{cases}\;\;$
Have to make homotopy: $(g\circ f)_{t}(x)=tx\;,t\in[0,1]$ but then if $t=0$ then $(g\circ f)_{t}(x)=0$ so every point in the ball goes to its center? On the other hand if $t=1$ then every point stays in its place.
Q1) Why $(g\circ f)(x)=0$? Should not $(g\circ f)(x)=x$ so that $g\circ f=id_{X}$?
Q2) does it mean now that $fg$ is homotopic to $id_{Y}$ and $g\circ f$ is homotopic to $id_{X}$? So that $X$ and $Y$ are homotopically equivalent? As far as I see it, this problem requires showing that $f\circ g$ is homotopic to $id_{Y}$ hence a homotopy function needs to be found between them, similarly $g\circ f$ is homotopic to $id_{X}$, again a homotopy needs to be determined, is the function $(g\circ f)_{t}(x)=0$ a homotopy function for both cases? This confuses me...
Thanks in advance.
To answer your first question, $gf(x)=g(\bullet)=0$ for all $x$. For your second question $fg$ is already equal to the identity, so you don't need a homotopy, although you can take the identity homotopy that doesn't do anything. $gf$ is indeed homotopic to the identity map on $X$ by the homotopy you gave $H(x,t)=tx$. I'm not really sure where your confusion lies.