Outside of Fomenko and Fuchs, I think it is more common to say that a pair has a homotopy extension property rather than say it is a Borsuk pair.

In any case, we call (X,A)(X,A) a Borsuk pair if the following problem can always be solved: given a homotopy ft:AYf_t : A\rightarrow Y and an initial condition F:XYF: X \rightarrow Y, we can extend these to a homotopy Ft:XYF_t : X \rightarrow Y which agrees with ftf_t and FF, namely:

F0=FF_0 = F
FtA=ftF_t|_{A} = f_t.

In other words, a Borsuk pair is one where we can extend homotopies. It turns out that we have the best case scenario:

Borsuk’s Theorem. Every CW pair is a Borsuk pair.

“Full glass retracts to the empty glass”, that’s how we fill in the glass.

Basically he’s covering section 5.5 of the book.

Now he’s doing the corollaries.

The point of defining the CW pairs and borsuk pairs like this and it doesn’t just hold for all pairs is that there may be a pathological counterexample, although he doesn’t have one on hand.

Cellular approximation of maps. Started by defining cellular maps, then made the cellular approximation theorem.

He first started proving the relative form of the theorem. He drew two lines for 0 and 1 corresponding to ff and fCWf^{\text{CW}}, and then noted that AA will just go to AA in both or something.

Free point lemma is so that we can get the free point. At each stage we need to use the free point lemma which he hasn’t proved yet.

Something about 7 or 5 balls but he thinks you’re able to get away with 3.

This Rm\R^m thing was we take the interior of this other shape and consider it as homeotopically equivalent to all of Rm\R^m.

Now proving the Free point lemma. Has some B0B_0 ball and preimage of it is K0=f1(B0)K_0 = f^{-1}(B_0). We can take two more “subballs” of B0B_0, B1B_1 and B2B_2, and their preimages are subshapes of K0K_0, called K1K_1 and K2K_2. We then use the Stone-Weierstrass Approximation Theorem t oapproximate it by polynomials. fK0pK0f|_{K_0} \approx p_{K_0} where fpK0ϵ\|f - p\|_{K_0} \le \epsilon

What he wants is to first of all improve the approximation so that it coincides with ff outside of K0K_0. Defined: f~=f+(pf)ρ=(1ρ)f+ρp\tilde{f} = f + (p-f)\rho = (1-\rho)f + \rho p where ρ\rho is basically our function defiend on the BBs but defined on the KKs? Now the map f~\tilde{f} sends: f~(K0K1)B2=\tilde{f}(K_0 - K_1) \cap B_2 = \varnothing because we can make the margin ϵ\epsilon exceeding the change made or something.

Now he used Sard’s Lemma to discuss that f~(K1)B2\tilde{f}(K_1) \cap B_2 had zero measure.

So he thinks three balls is enough.

Applications. dimXn\dim X \le n, YY, sknY=y0\text{sk}_nY = y_0, π(x,Y)=one-element=πb((X,x0),(Y,y0))\pi(x,Y) = \text{one-element} = \pi_b((X,x_0), (Y,y_0)).

YY is called nn-connected if SkYS^k \rightarrow Y is homotopic to a constant map knk\le n.

Examples:

Proposition. Any nn-connected CW-complex XX is homotopy equivalent to a CW-complex XX' such that sknX=x0\text{sk}_n X' = x_0.

Proof by induction on nn.

The argument is somewhat similar for both the base and the step, as long as you use some imagination. We can think of sk0X\text{sk}_0X as a boquet of zero-dimensional spheres. Path connected, so we can connect them with x0x_0. This path can be pushed to the 11-dimensional skeleton. Later inductively he said smth like the n1n-1-skeleton of XX is a boquet of n1n-1-dimensional spheres. Something about legitimately being an n+1n+1 dimensional cell because its boundary lies in the nn-dimensional skeleton.

Apparently the idea is pretty simple and the images are very transparant on its own. He was going to assign one of these proofs as homeworks but it’s very visual. This is very important for the course so you’ll bet I’m going to OH lol 💀.

There is some important idea of “pushing” but not sure.