1
$\begingroup$

It is known that every smooth manifold can be homotoped to a cell complex.

In particular this is true for manifolds with boundary.

My question: Under the homotopy to a cell complex, is the boundary homotoped to a co-dimension 1 sub-complex?

Assuming the above is true I claim:

If $M$ is an orientable compact 3-manifold with boundary such that there is defined $s\in\Gamma(\partial M,V_2(TM))$ a section of the 2-frame bundle of $TM$, then $s$ extends to $\tilde s\in\Gamma(M,V_2(TM))$

Proof:

Step 1

I define $\tilde s=s$ on $\partial M^{(1)}$, the 1-skeleton of $\partial M$.

Step 2

I extend $\tilde s$ over $M^{(1)}$, because $\pi_0(SO_3)=0$.

Step 3

I then extend it over $M^{(2)}$ (because $w_2(M)$ vanishes for any orientable compact 3-manifold, due to the Wu formula and the vanishing of $w_1(M)$ as $M$ is orientable)

Step 4

Now, having obtained $\tilde s$ in particular on $\partial M^{(2)}$, I can homotope $\tilde s$ to $s$ over $\partial M^{(2)}$, because they already agree on $\partial M^{(1)}$ and the obstruction to homotoping over the 2-skeleton is just $\pi_2(SO_3)$ which is the trivial group.

Step 5

Now, having obtained $\tilde s$ over $M^{(2)}$, we extend it to all of $M$, because again, the obstruction to extending over the 3-skeleton is the trivial group $\pi_2(SO_3)$.

I need this reviewed.

$\endgroup$
3
$\begingroup$

If $M$ is a torus with a hole removed (giving a boundary circle), the retraction will give a wedge of two circles, and the boundary circle of the torus maps nontrivially to this wedge. Therefore it does not map to a codimension 1 subcomplex.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.