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 $\tilde s\in\Gamma(M,V_2(TM))$


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.


1 Answer 1


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.


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