# Pushforward of a Vector Field by a Diffeomorphism

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.

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 $$F^{-1}$$.
• 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! Jun 23, 2020 at 0:05