Relative version of Whitney Immersion Theorem

Question

I asked the following question on StackExchange but received no response:

Let $M$ be a smooth $n$-dimensional compact manifold with boundary. Let $U$ be an open neighborhood of $\partial M$. Assume that we have a fixed immersion $\bar f : U \to \mathbb{R}^{2n-1}$. Can we always find an immersion $f : M \to \mathbb{R}^{2n-1}$ such that $f |_{\partial M} = \bar f |_{\partial M}$?

I believe this should work in $\mathbb{R}^{2n}$. Since $\bar f$ is an immersion, $d\bar f$ is a bundle monomorphism $TU \to U \times \mathbb{R}^{2n}$ covering $\bar f$. We can think of this as a section of the associated bundle over $U$ whose fibers are the Stiefel manifold $V_{n}(\mathbb{R}^{2n})$. Since $\pi_{n-1}\left(V_{n}(\mathbb{R}^{2n})\right) = 0$, we can extend the section from $\partial M$ to $M$. Now we can apply Smale-Hirsch to get the desired immersion $f : M \to \mathbb{R}^{2n}$.

Does the statement hold in $\mathbb{R}^{2n-1}$?

EDIT: To incorporate Oscar's comment, let's assume that we can perturb $\bar f$, if necessary. We can also assume that $U$ is connected.

Answer 1

No, you can't generally do this even with the added assumptions. The bundle $\tau = TS^{n-1} \to S^{n-1}$ is non-trivial for $n-1 \neq 1,3,7$, but it is stable after (one) trivialisation. Hence $\tau \oplus TD^n\vert_{S^{n-1}}$ is a trivial bundle over $S^{n-1}$.

By Smale--Hirsch this means there is an immersion $i: S^{n-1} \times [0,\epsilon] \hookrightarrow \mathbb{R}^{2n-1}$ whose ($(n-1)$-dimensional) normal bundle is isomorphic to (the pullback of) $\tau$. If $i$ extended to an immersion of $D^n$ then the normal bundle would extend to $D^n$ and hence be trivial, but it is not.