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?