Background:
This is from Livingston unplished notes in knot concordance:
Proposition 1.7.1 claims: Let $K$ be a knot in $S^3$. Then every Seifert surface $F$ for $K$ has a function $f:S^3-\nu(K)\to S^1$ such that $f$ is transverse to some point in the circle and its preimage is $F$. That is, there is $p$ in $S^1$ such that $F= f^{-1}(p)$.
He doesn't provide a proof, but a sketch. First, we can build $f$ by making it be a projection $S^1\times S^1\to S^1$ in $\partial\nu(K) =S^1\times S^1$ and thus the boundary of the Seifert Surface $F$ (so rigorously we are thinking of $F$ now bounding a longitude of $\nu(K)$) is sent to the same point in $S^1$.
Then we extend $f$ to be constant on $int(F)$, so $f$ now is extended to $\partial\nu(K)\cup F$.
Then he claims by obstruction theory we can extend to the remaining 3-cell.
I understand from obstruction theory that in a CW complex, if we have a function on the 2-skeleton to $S^1$, we can extend it to the 3-skeleton as well. The obstruction lives in $H^3$ with $\pi_2(S^1)=0$ coefficients, so it vanishes.
Question:
The torus has the classic CW decomposition in a square. There's NO CW decomposition of the torus with its 1-skeleton being a single 0-cell and a single 1-cell. We need to add at least one more 1-cell so as to possibly have it be some 1-skeleton of some CW decomposition.
For us to use this obstruction-theoretic fact above, we need to guarantee that there is a CW decomposition of $S^3-\nu(K)$ such that $\partial\nu(K)\cup F$, where we extended $f$ to, is its 2-skeleton. Or at least, that we can extend $\partial\nu(K)\cup F$ to a few more 2-dimensional simplices so as to comprise a 2-skeleton of some CW decomposition of $S^3-\nu(K)$!
How do we see this is the case for $\partial\nu(K)\cup F$?