Do we really need to use the assumption to show that $\phi$ is well defined. (Serge Lang Calculus of Several Variables)

Question

I am reading "Calculus of Several Variables 3rd Edition" by Serge Lang.

Do we really need to use the assumption to show that $\phi$ is well defined.

Answer 1

Olivier gives the answer in his comment, but you seem to have missed the point. It is true that $\phi$ is well-defined for all $X\neq0$ not on the circle, because Lang specifies a unique path from $(1,0)$ to $X$. And obviously $\phi((1,0))=0$.

But consider points of the form $X_\epsilon=(\cos\theta,\sin\theta)$ for $\theta=2\pi-\epsilon$, where $0<\epsilon<2\pi$. We'd like $\phi$ to be differentiable, so we need it to be continuous; hence $\phi(X_\epsilon)$ should approach $\phi((1,0))=0$ as $\epsilon\to0$. The assumption guarantees this.