Prove that the Brouwer theorem is false for the open ball $|x|^2 Prove that the Brouwer theorem is false for the open ball $|x|^2 <a$
Brouwer Theorem: Any smooth map $f$ of the close unit ball $B^n \subset R^n$ tin to it self must have a fixed point.
I need to find a counter example, so I let $f: B^k \to B^k$ be a composition function 
$$B^k \to^g R^k \to^h R^k \to^{g^{-1}} B^k$$
where $h$ be define as $h(x) =x+a$ for $a \not =0$ . Note that $h: R^k \to R^k$ has no fixed point, so $f: B^k \to B^k$ doesn't have any fixed point either.
Is my reasoning acceptable? or it need more work? If so, please help me improve it.

As has been established in the comments, your construction works because $f$ has a fixed point if and only if $h$ does, but $h$ was chosen so that it doesn't have a fixed point.
Another map that works is $f : B(0, a) \to B(0, a)$ given by $f(x) = \frac{1}{2}(x+ae_1)$ where $e_1 = (1, 0, \dots, 0)$. This sends a point $x \in B(0, a)$ to the midpoint of the line segment between $x$ and $ae_1 \in \partial B(0, a)$. It can be verified either algebraically or geometrically that $f$ has no fixed point. If the map were extended to $\overline{B(0, a)}$, it would have a fixed point at $ae_1$.
