# Winding number of a curve contained in a triangle around a point outside the triangle

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?

• Homotopies that don‘t pass through $a$ preserve the winding number. Dec 14, 2019 at 19:59

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 $$[\gamma] \subseteq \Delta.$$