Let $C_1$ and $C_2$ be two great circles in $S^2$, intersecting at the points $p,q$. If we consider $C_1$ and $C_2$ as curves starting and ending at $p$. Prove that $C_1$ and $C_2$ are homotopic fixing endpoints. So how to prove that?
What I have done so far is:
1st method: As every great circle is the intersection of a plane through origin & the sphere $S^2$. So, we can deal with the planes only. Let two planes passing through origin be $a_1x+b_1y+c_1z=0$ and $a_2x+b_2y+c_2z=0$. Now the combination $H((x,y,z),t)=[(ta_1+(1-t)a_2)x+(tb_1+(1-t)b_2)y+(tc_1+(1-t)c_2)z] \cap S^2$ as the homotopy function from $S^2 \times I \to S^2$. But does the map preserve the end point? If not then modify it & show that it will solve our purpose.
2nd method: I can rotate the axis & take it on $C_1$ s.t it works as $X-Y$ plane then the other circle(or plane) be at the angle $\phi$ with plane $X-Y$ in the $(r,\theta,\phi)$ co-ordinate. So, we can give the homotopy $H:S^2 \times I \to S^2$ s.t $(\phi,t)\mapsto t\phi$. I know that this map will preserve the end points but is this map is written in a proper way and give me the funtion of rotation of axis which is stated at the begining of 2nd method.
Any other illustrated solutions are also welcome. Illustrated because I am new in algebraic topology.
Either method works. They essentially the same homotopy, just different expressions: The angle between $\vec{n}_t$ and $\vec{n}_1$ is the angle in the second homotopy.
They fix the endpoint $p$ because all planes $H_t$ pass $p$ and $q$.
Note: You may need to pay attention to the orientation of the paths $C_1$ and $C_2$, as the above homotopy may send $C_1$ to $-C_2$. In the picture, what if we choose $-\vec{n}_2=(-a_2,-b_2,-c_2)$ for a normal to $H_1$?