I am considering the following problem:

If a closed curve $\gamma$ is contained in a triangle $\Delta$ and $a$ is a point outside of $\Delta$ then $I(\gamma, a) = 0$, where $I$ is the winding number of $\gamma$ around $a$.

Intuitively this makes sense, $\Delta$ is simply connected, and so since $a$ is outside $\Delta$, no curve in $\Delta$ can wind around $a$.

However, the solution I am given to this problem makes the following argument: By convexity of $\Delta$, for all $p \in \Delta$, the straight line homotopy from $\gamma$ to $p$ is contained in $\Delta$, and since $a \not \in \Delta$, defining $\Gamma_0(t) = \gamma(t)$ and $\Gamma_1(t) = p$ gives $I(\Gamma_0, a) = I(\Gamma_1, a) = 0$.

I do not understand how we can conclude that $I(\gamma, a) = 0$ from the fact that $I(\Gamma_1, a) = 0$. Is there some obvious fact about homotopies I am missing?

  • 3
    $\begingroup$ Homotopies that don‘t pass through $a$ preserve the winding number. $\endgroup$
    – bsbb4
    Dec 14, 2019 at 19:59

1 Answer 1


This bypasses part of your question, but because it is nearly a one-liner, it should be mentioned:

The closed set $\Delta$ is the intersection of three closed half-planes. As $a \notin \Delta,$ at least one of them doesn't contain $a.$ Any continuous choice of $\arg(z - a)$ for $z$ in such a half-plane is also valid for $z \in \Delta,$ so $I(\gamma, a) = 0$ if $[\gamma] \subseteq \Delta.$


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