Vector field tangent to a submanifold and transverse to the zero section

Question

In Hirsch's Differential Topology there's the following :

Suppose a compact $n$-manifold can be expressed as $A\cup B$ where $A,B$ are compact $n$-dimensional submanifolds and $A\cap B$ is an $(n-1)$-dimensional submanifold. Then $\chi(A\cup B)=\chi(A)+\chi(B)-\chi(A\cap B)$.

Trying to solve this a question the following came to my mind :

Suppose we have a submanifold $N$ of $M$.Is it possible to have a section $f:M\rightarrow TM$ such that $f|_N\in TN$, where we make the canonical identifications, and such that $f$ is transverse to the zero section ?

Now I belive we can construct a vector field $f$ in $M$ that is tangent to $N$ using local charts but for the added requirement of it being transverse to the zero section I am not sure how it could be done, sure we can use the transversality theorem to get maps $h_k\rightarrow f$ that are transverse to the zero section , but no matter how close I can approximate $f$ I can always ruin the fact that is tangent to $N$, not sure if there is anymore conditions I can put to stop this from hapenning.

Does anyone have any thoughts on this ? Thanks in advance.

Answer 1

Choose an open cover of $M$ where, for each open set $U$ in the cover that intersects $N$, there is a chart $U \subseteq \mathbb R^n$ that sends $N$ to $\mathbb R^{n-1}$.

Choose a partition of unity for this open cover.

For each open set, define a vector field on $\mathbb R^n$ as $\sum_i f_i \frac{d}{dx_i}$ where $f_i$ is a random smooth function for $i =1,\dots,n-1$ and $f_n$ is a random smooth function times $x_n$. (In fact I think it suffices to take random affine functions.) Then combine these vector fields into a vector field on $M$ using the partition of unity. By construction, its restriction to $N$ liles in the tangent bundle of $N$.

Now we need to check it's transverse. At each point of $M$ not in $N$, the vector field vanishing at that point is a codimension $n$ condition and vanishing non-transversely is codimension $n+1$, so overall the vector field vanishing non-transversely is codimension $1$. Similarly at each point of $N$, vanishing is codimension $n-1$ and vanishing non-transversely is codimension $n$, so overall the vector field vanishing non-transversely is codimension $1$, so a random/generic choice works.

Alternatively, we could construct a vector bundle $T'M$ with a map to $TM$ that is an isomorphism away from $N$ but over $N$ has image $TN$. In our open chart $U$, this map would be given by a diagonal matrix with entries $1,\dots,1,x_n)$, and one can check that this glues between different open charts. Then finding a section of this vector bundle transversal to the inverse image of the zero-section using the standard transversality theorem should work.