Notes for AKT-140108/0:24:50: Difference between revisions
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I am still not clear on why the image of the unlink is the equator of the sphere. I am wandering why not any other circle above or below the equator. Do we homotop the image to the equator? |
I am still not clear on why the image of the unlink is the equator of the sphere. I am wandering why not any other circle above or below the equator. Do we homotop the image to the equator? |
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Since we have assumed that the embedding is invariant, we can keep the two circles in the same plane. Then, the directional vector would always be inside of the plane. Since we are free to choose which plane we want the unlink circles to be, without loss of generality, the two circles are chosen to be in the xy-plane. Then the directional vector will not have the z-component, hence the image of the map can only be on the equator (if we choose some other planes, the image would be a great circle on <math>S^2</math> by rotating the equator). |
Latest revision as of 22:08, 30 May 2018
I am still not clear on why the image of the unlink is the equator of the sphere. I am wandering why not any other circle above or below the equator. Do we homotop the image to the equator?
Since we have assumed that the embedding is invariant, we can keep the two circles in the same plane. Then, the directional vector would always be inside of the plane. Since we are free to choose which plane we want the unlink circles to be, without loss of generality, the two circles are chosen to be in the xy-plane. Then the directional vector will not have the z-component, hence the image of the map can only be on the equator (if we choose some other planes, the image would be a great circle on by rotating the equator).