Can we consider a "weak bundle" as a topological vector bundle?

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Let $S$ be a smooth compact $k$-dimensional manifold.

Suppose we are given a $W^{2,2}$ immersion $F:S \to \mathbb{R}^d$, that is $F \in W^{2,2}(S,\mathbb{R}^d)$ and $dF$ is an immersion a.e.

Is there any sense in considering $dF(TS)$ as a topological vector bundle over $S$?

(we endow $dF(TS)$ with the subspace topology induced by $T\mathbb{R}^d$).

The point is that while the fibers $p \to T_pS$ change continuously (even smoothly), $p \to dF_p$ does not...

I wonder if there is anything intelligible one can say of this "weak bundle" (whose fibers are defined only almost everywhere).

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Probably there is no reasonable way to consider this weak bundle as something similar to a true smooth bundle. I guess this is part of the reason why, when phrasing regularity theory (e.g weak harmonicity), we usually work directly with the smooth "ambient bundle" $T\mathbb{R}^d|_{S}$, and not with $dF(TS) \subseteq T\mathbb{R}^d|_{S}$.

From aesthetic point of view, I would prefer a more intrinsic version of regularity theory, but maybe this is too much to ask for.

Admittedly, I did not write here anything mindblowing, but I thought some answer is needed, so we could take away this question from the "unanswered list".