Let $f:\overline{B}_r(0)\rightarrow \overline{B}_r(0)\setminus\{0\}$ be a continuous function. Suppose that $f(x)=x$, when $x\in \partial \overline{B}_r(0)$. Then we can consider a continuous function $g:\overline{B}_r(0)\rightarrow \partial\overline{B}_r(0)$ $$g(x)=r\frac{f(x)}{|f(x)|}.$$
Clearly $g(x)=x$, when $x\in \partial \overline{B}_r(0)$, and this is in conflict with the no-retraction theorem .
Question:
Let $f:\overline{B}_r(p)\rightarrow \overline{B}_r(0)\setminus\{p\}$ be a continuous function. Suppose that $f(x)=x$, when $x\in \partial \overline{B}_r(p)$. Then we can consider a continuous function $g:\overline{B}_r(p)\rightarrow \partial\overline{B}_r(p)$ $$g(x)=r\frac{f(x)-p}{|f(x)-p|}+p.$$
Now it is not at all clear to me that we can conclude a contradiction. Is there any? What might be a remedy or an explanation?
Given the function $f:\overline{B}_r(p)\to\overline{B}_r(p)\setminus\{p\}$, let
$$g:\overline{B}_r(0)\to\overline{B}_r(0)\setminus\{0\}:x\mapsto f(x+p)-p\;,$$
and apply the original result to $g$.