Room for Daminark and Fargle

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Fargle

23:03

nerd

Daminark

Lmao

Truly

So the first thing to understand is that $S^n = I^n/\partial I^n$

Fargle

Makes perfect sense.

Daminark

Yeah, though I do want to go through this carefully just to make sure

In the case of $S^1$ this is definitely clear

Like, you take $[0,1]$ and glue $0$ to $1$

In the case of $S^2$, we're taking a square and turning the edge into a point, that's a bit trickier to visualize

Fargle

Yeah but I have that down.

You can also take the 2D disk and identify its top semjcircumference with its bottom semicircumference, and then take that down to a point.

Ends up being the same since $D^2 \cong I^2$.

Daminark

Oh wait that'd be convenient

Is this true in general? If I see that picture I'm much more alright with it

Right I believe it actually

Because you can just place $D^n$ sorta in the middle of $I^n$ and then scale

Fargle

23:16

I don't know if it's true in general, but I know it works for $S^1$ and $S^2$ (where for $S^1$, you just glue together the two connected components of $D^1$'s boundary)

it ends up in that case exactly equivalent to the quotient construction you give for $S^n$

Daminark

So with $D^n$ I know you can sorta press it down and make it the top hemisphere of $D^{n+1}$

(Including the equator)

Fargle

@Daminark Careful--of $S^n$.

Daminark

By press it down I mean just curve it until it becomes a hemisphere

No but think about it

Fargle

@Daminark I agree with you, but it only becomes the surface of the thing.

Daminark

Oh, right yeah

Fargle

23:19

But you're right on the money, realizing $S^n = \partial D^{n+1}$.

Daminark

But then what happens it that you now sorta stretch it over $S^n$ until the equator becomes a pole

That's taking it mod the boundary

Fargle

Yeah, it's a sort of stereographic thing.

Daminark

But then boom you have $S^n$

Now I'm happy

Fargle

Alright, so ATop.

I'm with you on the quotient construction of $S^n$.

Daminark

So now we can understand higher homotopy groups as maps from $I^n$ to $X$ but so that if any point has a coordinate of $0$ or $1$, it's mapped to the basepoint $x_0\in X$

Fargle

23:26

Alright, let me picture this in the second case.

So you're mapping the unit square into the space continuously, so that any point on the boundary of the unit square is mapped to a certain basepoint.

Daminark

Yup

Fargle

Which essentially means you're finding the "copies" of $S^2$ that exist in the space and are equivalent mod homotopy.

Daminark

I'm basically making the boundary of the box into a basepoint (remember it gets crushed into a point when we're thinking about it as $S^n$)

Yeah

Fargle

And then for the third homotopy group, you're finding the "copies" of $S^3$ set at a basepoint that are equivalent mod homotopy, and so on.

Daminark

Yup

Fargle

23:29

So the identity is clearly just the contraction of $I^n$ to the basepoint.

Daminark

Exactly

Fargle

The operation of concatenation is clearly closed, since $I^n$ is homeomorphic to two copies of itself set side-by-side.

Associativity is also obvious.

Daminark

The reason I want to look at it like this is because the group operation is now clear

You traverse over coordinates running from $0$ to $1$

I think the idea would be running the first coordinate, then the second, etc, right?

Fargle

Right, analogous to the fundamental group.

And then the operation involves running over the first map from $0$ to $1/2$ in each coordinate, and then the second map from $1/2$ to $1$ in each coordinate.

The inverse of a map is just running over those coordinates in reverse.

Daminark

Sounds good

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