Prove that $S_R^+(b)$ and $S_r^+(a)$ are homotopic in a domain $D$.

Question

currently I'm working on the following exercise:

Let $D \subset \mathbb C$ be a domain. Let $a,b \in \mathbb C$ and $r,R > 0$ such that $B_r(a) \subset B_R(b)$ and \begin{align*} A := \{z \in \mathbb C: |z-b| \le R, \, |z-a|\ge r\} \subset D. \end{align*} Prove that the circle $S_R^+(b)$ and $S_r^+(a)$ are homotopic in $D$.

I wanted to use the following homotopy: \begin{align*} H(t,s) &:= (1-s)(b+Re^{it}) + s(a+re^{it}). \end{align*} We have that \begin{align*} H(t,0) = b+Re^{it} = S_R^+(b), \quad H(t,1) = a+re^{it} = S_r^+(a), \quad \forall t \in [0,2\pi]. \end{align*} Furthermore \begin{align*} H(0,s) = (1-s)(b+R)+s(a+r) = H(2\pi,s), \quad \forall s \in [0,1]. \end{align*} We need to show that $H(t,s) \in D$, $\forall (t,s) \in [0,2\pi] \times [0,1]$. We can do that by showing $H(t,s) \in A$. We have that \begin{align*} |H(t,s)-b| &= |(1-s)(b+Re^{it}) + s(a+re^{it})-b| \\ &= |(1-s)b-b+sa+(1-s)Re^{it}+sre^{it}| \\ &= |b(1-s-1)+sa+e^{it}((1-s)R+sr)| = |sa-sb+e^{it}(R-sR+sr)| \\ &= |s(a-b)+e^{it}(R-s(R-r))| \le |s||a-b|+|R-s(R-r)|. \end{align*} We know that $B_r(a) \subset B_R(b)$. That means \begin{align*} |z-a| < r \Rightarrow |z-b|<R, \quad \forall z \in B_r(a). \end{align*}

But I don't know how to use this in my estimation.

Edit: I worked my way through the solution which I got, and it uses the estimation $|a-b| \le R-r$. I don't understand why this is true.

Answer 1

Some comments:

• Good work. Nevertheless, I don't know why you need your "Furthermore...". :-?
• Also, in order to show that $H(s,t) \in A$, you need to prove that $|H(s,t) - a| \geq r$ too, don't you?
• As for your first question, I would try the following estimation instead:

\begin{align*} |H(t,s)-b| &= |(1-s)(b+Re^{it}) + s(a+re^{it})-b| \\ &= |(1-s)(b+Re^{it}) +sb -sb + s(a+re^{it})-b| \\ &= |(1-s)Re^{it} + s(a-b + re^{it})| \\ &\le (1-s)|Re^{it}| + s|a-b| + s|re^{it}| \\ & = (1-s)R + s|a-b| + sr \\ \end{align*}

Now you use the hint $|a-b| < R - r$, and get

\begin{align*} &< (1-s)R + s(R-r) + sr = R \end{align*}

So, it's ok: your homotopy $H(s,t)$ never goes outside the big disk $B_R(b)$.

• As for your second question, namely this last inequality $|a-b| < R - r$ comes from the fact that $a\in B_R(b)$, so $|a-b| < R$, plus the fact you can still go further away $r$ steps in the direction of the vector $a-b$ and you're still inside the big disk $B_R(b)$, because $B_r(a) \subset B_R(b)$. Hence, $|a-b|+r < R$.