First, let us see what exactly does not work for $S^{n}$ , $n>1$ .

The proof that there is no retraction of $B^{2}$ onto $S^{1}$ was based on the fact that otherwise $j_{*}$ must be injective, but the fundamental group of $S^{1}$ is not trivial, while the fundamental group of $B^{2}$ is trivial. This does not work even for $\mathbb{R}^{3}$ . The fundamental groups of $S^{2}$ and $B^{3}$ are both trivial, because every loop can be contracted to a point. In fact, it is easy to see that to show that $\mathbb{R}^{3}$ and $\mathbb{R}^{3}-0$ are different it is not enough to consider just loops. Intuitively, every loop in $\mathbb{R}^{3}$ minus finite number of points can be continuously transformed to a point (avoiding the removed points). So, to show that the spaces are different we would need to consider something more than loops. A loop is a continuous function of $S^{1}$ into the space (induced by the path via the quotient $I\rightarrow S^{1}$ ). Intuitively, if we remove a point from $\mathbb{R}^{3}$ , then we can still transform any circle to a point (using the third dimension to avoid the removed point if necessary), however, if we consider continuous maps of the sphere $S^{2}$ into $\mathbb{R}^{3}$ , then not every such a continuous function can be transformed to a point, in particular, if the removed point is inside the initial image of $S^{2}$ , then such a function cannot be continuously transformed to a point. But if there is no removed point, then every function from $S^{2}$ into $\mathbb{R}^{3}$ can be continuously transformed to a point. This shows that while “the higher order fundamental group” of $\mathbb{R}^{3}$ is still trivial, the one of $\mathbb{R}^{3}-0$ is not, and we could use that fact to show that there is no retraction, but we just did not define such notions. So, again, considering just paths is not enough to distinguish $\mathbb{R}^{3}$ and $\mathbb{R}^{3}-0$ , or to show there is no retraction of $B^{3}$ onto $S^{2}$ .

Lemma 55.3 (1) and (2) tells us that if a continuous $h:S^{1}\rightarrow X$ is nulhomotopic then $h$ continuously extends to $B^{2}$ . The proof of this part seems to work for higher dimensions as well. Moreover, if we can extend $h$ to $k$ on $B^{n+1}$ , then the composition of the quotient $S^{n}\times I\rightarrow B^{n+1}$ and $k$ gives a homotopy between $h$ and a constant map, so that it is “iff” in any dimension.

Then, to show that the inclusion map $j:S^{1}\rightarrow\mathbb{R}^{2}-0\mbox{ or }S^{1}$ is not nulhomotopic, Lemma 55.3 (3) was used: there is a retraction, so that $j_{*}$ is injective and, hence, nontrivial, therefore, $j$ is not nulhomotopic. Regardless, of whether Lemma 55.3 (3) is equivalent to $h$ being nulhomotopic or not in higher dimensions, we cannot use it here, because $S^{n}$ , $n\ge2$ , is trivial, so that even if $j_{*}$ is injective, its image does not have to be a nontrivial group. This shows that we cannot use this approach for (a) and (b). However, we can still use the “no retraction” fact and Lemma 55.3 (1) and (2) proved for higher dimensions.

Theorem 55.5 showing that non-vanishing fields have inward/outward vectors on the border, used the fact that the restriction of the field to the border (as a function into $\mathbb{R}^{2}-0$ ) is both extendable, hence, nulhomotopic, and also homotopic to the inclusion map if there is no inward vectors, hence, in that case the inclusion map must be nulhomotopic. Once we prove (b), this approach would work for (c).

The proof of Theorem 55.6 (Brouwer) was based on Theorem 55.5, so that (d) follows from (c) in the same way.

Then, Corollary 55.7 can be proved the same way using the higher dimensions Brouwer theorem, which gives us (e).

Finally, our proof of Exercise 2 is based on the fact that if $h$ is extendable, then it has a fixed point, and the approach works for any dimension, proving (f).

To sum up, we cannot directly prove that there is no retraction of $B^{n+1}$ onto $S^{n}$ for $n\ge2$ , so this is given. However, we can still prove the equivalence of Lemma 55.3 (1) and (2) for any dimension, and use this advanced lemma to prove (a) and (b). From this three facts (the advanced lemma, (a) and (b)) everything else follows in almost exactly way as in the Section 55.

(a) If $i$ were nulhomotopic, it would be extendable onto $B^{n+1}$ , and the extension would be a retract of $B^{n+1}$ onto $S^{n}$ .

(b) If $j$ were nulhomotopic, it would be extendable to $k:B^{n+1}\rightarrow\mathbb{R}^{n+1}-0$ , there is a retract $h:\mathbb{R}^{n+1}-0\rightarrow S^{n}$ given by $h(x)=x/||x||$ , so that $h\circ k$ is a retract of $B^{n+1}$ onto $S^{n}$ .

(c) The proof of Theorem 55.5 works almost exactly, substituting (b) for Corollary 55.4.

(d) The proof of Theorem 55.6 works as well using (c), in other words, $f(x)-x$ has no outward pointing vectors on the boundary, hence, it is not nonvanishing.

(e) The proof of Corollary 55.7 works as well using (d), in other words, $Ax/||Ax||$ is well-defined and continuous on the non-negative part of $S^{n}$ , hence, has a fixed point.

(f) We repeat our proof of Exercise 2 here: if $h$ is nulhomotopic, then it is continuous and extendable to a continuous function $k:B^{n+1}\rightarrow S^{n}$ , then $j\circ k$ has a fixed point $x$ (using (d)) in $j(S^{n})=S^{n}$ , so that $x=j\circ k(x)=j\circ h(x)=h(x)$ . Similarly, if $g:S^{n}\rightarrow S^{n}$ is given by $g(x)=-x$ , then $g\circ h$ is nulhomotopic, and for some point $x=g\circ h(x)=-h(x)$ .