I think a couple of things would be helpful here: 1. At 2:30, explain more clearly that you're talking about deforming the loop *within S1*, i.e., you can't just deform the entire circle into a point because this requires moving off the circle. 2. At 5:20 or at the end, explain why the function phi doesn't work if f has a fixed point.
Yes, you are right, and I do admit the video is pretty informal, although this choice is mainly to allow the video to be more accessible. Here is a brief answer to your points for other interested viewers: 1. Here by "continuous deformation of a loop in S^1", I mean the existence of a continuous map [0,1] x S^1 \to S^1 (i.e., a homotopy). In particular, at each fixed t \in [0, 1], we get a map S^1 \to S^1. I believe I was also mainly working with basepoint-preserving homotopy in the video, although this is not necessary for the argument. For your correction, I mention that it's not quite accurate to state that "you can't just deform the entire circle into a point because this requires moving off the circle": Instead, the argument is that (intuitively) there is no deformation that continuously contracts a loop into a constant loop. 2. Note that we cannot even really draw a well-defined ray from p to f(p) if p and f(p) are not distinct points.
I think you meant to say at 6:20 that "if p is on S^1, then f(p) is somewhere else. But the vector starting at f(p) pointing toward p still intersects S^1 at p".
Not quite: I believe what is stated in the video is precise. The version you suggested is inaccurate due to two reasons: 1) We really should work with a ray, not a vector, and 2) We cannot consider the ray starting at f(p) but must consider the ray starting at p because the possibility that f(p) lies on S^1 can cause continuity problems.
@@LetsSolveMathProblems I think they meant what you did mean (but worded less formally), pointing to different thing: that in the video it was "ray starting at p and in the direction of this ϕ(p) to p vector" and not "f(p) to p vector"
Probably not, as I am honestly not familiar with a proof of Schauder fixed-point theorem. Personally, the way I have seen the Brouwer's fixed point theorem proven in higher dimensions is by using homology/higher homotopy groups (see Hatcher's "Algebraic Topology") or smooth manifold theory (see Milnor's "Topology from the Differentiable Viewpoint").
I think a couple of things would be helpful here:
1. At 2:30, explain more clearly that you're talking about deforming the loop *within S1*, i.e., you can't just deform the entire circle into a point because this requires moving off the circle.
2. At 5:20 or at the end, explain why the function phi doesn't work if f has a fixed point.
Yes, you are right, and I do admit the video is pretty informal, although this choice is mainly to allow the video to be more accessible. Here is a brief answer to your points for other interested viewers:
1. Here by "continuous deformation of a loop in S^1", I mean the existence of a continuous map [0,1] x S^1 \to S^1 (i.e., a homotopy). In particular, at each fixed t \in [0, 1], we get a map S^1 \to S^1. I believe I was also mainly working with basepoint-preserving homotopy in the video, although this is not necessary for the argument. For your correction, I mention that it's not quite accurate to state that "you can't just deform the entire circle into a point because this requires moving off the circle": Instead, the argument is that (intuitively) there is no deformation that continuously contracts a loop into a constant loop.
2. Note that we cannot even really draw a well-defined ray from p to f(p) if p and f(p) are not distinct points.
I think you meant to say at 6:20 that "if p is on S^1, then f(p) is somewhere else. But the vector starting at f(p) pointing toward p still intersects S^1 at p".
Not quite: I believe what is stated in the video is precise. The version you suggested is inaccurate due to two reasons: 1) We really should work with a ray, not a vector, and 2) We cannot consider the ray starting at f(p) but must consider the ray starting at p because the possibility that f(p) lies on S^1 can cause continuity problems.
@@LetsSolveMathProblems I think they meant what you did mean (but worded less formally), pointing to different thing: that in the video it was "ray starting at p and in the direction of this ϕ(p) to p vector" and not "f(p) to p vector"
Ah yes, you and the previous commenter are absolutely right. I should have written f(p) (not phi(p)) at 6:22. Thank you for pointing this out!
Can you go for schauder fixed point theorems then again brouwer in R^n
Probably not, as I am honestly not familiar with a proof of Schauder fixed-point theorem. Personally, the way I have seen the Brouwer's fixed point theorem proven in higher dimensions is by using homology/higher homotopy groups (see Hatcher's "Algebraic Topology") or smooth manifold theory (see Milnor's "Topology from the Differentiable Viewpoint").