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A question concerning when the pushforward of a vector field is well-defined. if $ F:N \rightarrow M $ is a smooth map between manifolds, the pushforward of a tangent vector $ X_p \in T_PN $ is given by $ F_{*,p}:T_pN \rightarrow T_{F(p)}M $, where $ \big(F_{*,p}(X_p)\big)f := X_p(f \circ F) $ for some $ f \in C^\infty_{F(p)}(M) $. However, (I think?) I understand that the pushforward of a smooth section $ X:M \rightarrow TM $ on the bundle $ (TM,M,\pi) $ doesn't generally exist, as the pointwise definition of the bundle map $ F_*:TN \rightarrow TM $, such that $ (F_*X)_{F(p)} = F_{*,p}(X_p) $ is ambiguously defined if $ F $ isn't one-to-one.

That being said, the references I've checked all uniformly state that a necessary and sufficient condition for this pushforward to exist is that $ F $ is a diffeomorphism, but this seems like overkill to me. If $ F $ is smooth and a bijection, isn't a smooth homeomorphism a sufficient condition? Furthermore, does $ F $ even need to be onto, as long as we don't care about points in $ M $ outside of the image of $ F $? Does $ F $ only need to be smooth and one-to-one, and that's it? I guess I don't see why we need $ F^{-1} $ to be smooth to pull this off. I'd normally assume I'm being too picky, but most proofs concerning such vector fields seem to begin with something like "Let $ F $ be a diffeomorphism...", so I assume I'm missing something critical.

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Suppose that $F$ were a smooth bijection $N\to U\subset M$, where $U$ is some open subset of $M$. Then for a (smooth) vector field $X\colon N\to TN$, we can define a map $F_*X\colon U\to TU$ by $$ F_*X\colon q\mapsto(F_*X)_q = dF_{F^{-1}(q)}(X_{F^{-1}(q)}). $$ This is well-defined as a set map from our assumption that $F$ be bijective onto its image. From the explicit formula $F_*X = dF\circ X\circ F^{-1}$, where $dF\colon TN \to TU$ is the bundle map induced by $F$, we see that $F_*X$ is continuous iff $F^{-1}$ is continuous (i.e., $F$ is a homeo onto its image), and that $F_*X$ smooth iff we suppose further that $F^{-1}$ is smooth (i.e. $F$ is a diffeo onto its image). So we see that the regularity of $F_*X$ is equivalent to the regularity of $F^{-1}$.

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    $\begingroup$ Ah, noting that $ F_*X = dF \circ X \circ F^{-1} $ is what I was missing. The subtle change of recasting the problem in terms of $ p = F^{-1}(q) $ immediately shines the light on the issue in my mind. Thanks for your help! $\endgroup$
    – Jerome
    Jun 23, 2020 at 0:05

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