Pushforward of a smooth map that is not injective

Question

Given a smooth map $\psi :M\rightarrow N$, we say the vector field $Y$ on $N$ is $\psi$-related to the vector field $X$ on $M$ if for all $x\in M$: $$d\psi_x(X_x) = Y_{\psi(x)} $$

Given a diffeomorphism $\phi :M\rightarrow N$, we can define the pushforward of the vector field $X$ on $M$ as the unique vector field $Y$ on $N$ that is $\phi$-related to $X$: $$\phi_*X = Y $$ is such that for all $x \in M$ $$d\phi_x(X) = Y_{\phi(x)} $$

The pushforward is defined more generally for smooth maps $\psi:M\rightarrow N$ in which case it is possible for more than one vector field to be $\psi$-related to $X$ and $\psi_*X$ is not defined.

This seems obvious to me if $\psi$ is not surjective in which case $Y$ could take on any value outside the image of $\psi$.

According to Wikipedia, "A more general situation arises when $\psi$ is surjective. Then a vector field X on M is said to be projectable if for all $y$ in $N$, $d\psi_x(X_x)$ is independent of the choice of $x$ in $\psi^{−1}({y})$."

It makes sense to me that in the above situation we can define $\psi_*X$.

Where my confusion lies is that it seems to me that the definition of $\psi$-relatedness already requires $d\psi_x(X_x)$ to be independent of the choice of $x$ in $\psi^{−1}({y})$: I would've said that was a condition of $d\phi_x(X_x) = Y_{\phi(x)} $ for all $x$

How is it then, that when $\psi$ is surjective but not injective, that several vector fields are $\psi$-related to $X$ ? How is the pushforward defined for non-injective maps ?

My best guess would be to say that $Y$ is $\psi$-related to $X$ if for all $y \in $ Im($\psi$) there exists $x\in M$ such that $$d\phi_x(X_x) = Y_{\phi(x)} $$

Is this on the right track ?

Answer 1

Let's say you are given a smooth map $\psi \colon M \rightarrow N$ and a vector field $X$ on $M$. Assume there exists some vector field $Y$ on $N$ that is $\psi$-related to $X$. Then we must have: $$ d\psi_{m} \left( X|_{m} \right) = Y(\psi(m)) $$ for all $m \in M$. This tells you two things:

1. A neccesary condition for the existence of $Y$ is that for all $n \in N$ and $m,m' \in \psi^{-1}(n)$ we must have $d\psi_{m} \left( X|_{m} \right) = d\psi_{m'} \left( X|_{m'} \right)$. This can indeed be rephrased as $d\psi_{m} \left( X|_{m} \right)$ is independent on $m \in \psi^{-1}(n)$.
2. Any $\psi$-related vector field $Y$ to $X$ (if it exists) is determined uniquely on $\operatorname{im}(\psi)$.

In particular, if $\psi$ is surjective there can be at most one $\psi$-related vector field to $X$ and if condition $(1)$ is satisfied, you can try to define $Y(\psi(m)) = d\psi_m \left( X|_{m} \right)$ for all $m \in M$. This is well-defined and if it is actually a smooth vector field, this will be the only $\psi$-related vector field to $X$. However, this doesn't have to be the case!

Consider for example $\psi \colon \mathbb{R} \rightarrow \mathbb{R}$ given by $\psi(x) = x^3$ and $X = \partial_{x}$. Then $\psi$ is smooth and bijective (in particular, surjective) and $$ d\psi|_{x} \left( \partial_x \right) = 3x^2 \partial_x|_{x^3} $$ so if you try and define $$ Y \left( x^3 \right) = 3x^2 \partial_x |_{x^3} \iff Y(x) = 3x^{\frac{2}{3}} \partial_x $$ you see that $Y$ is not smooth. So there isn't any smooth $\psi$-related vector field to $X$.

This situation does cannot arrise if you add further regularity assumptions on $\psi$. For example, if $\psi$ is a submersion then this cannot happen.

Answer 2

The pushforward of non-injective maps is defined precisely by projectability.

Let $M$ and $N$ be smooth manifolds and $M\xrightarrow{\phi}N$ a smooth map. Suppose $\phi$ is surjective and not necessarily injective.

Consider $y\in N$ and the following set, \begin{equation*} \{x_1,\ldots,x_n\}=\phi^{-1}(y), \end{equation*} where $x_i\neq x_j$ for $i\neq j$.

If $X$ is a vector field on $M$, then it certainly may be the case that \begin{equation*} X_{x_i}\neq X_{x_j},\;i\neq j. \end{equation*} It may even be the case that \begin{equation*} d\phi_{x_i}(X_{x_i})\neq d\phi_{x_j}(X_{x_j}),\;i\neq j. \end{equation*} This would be a case when we can have several vector fields $\phi$-related to $X$. If we cannot pick a unique representative, then different choices woud lead to different pushforwards.

However, if we have the condition that \begin{equation*} d\phi_{x_i}(X_{x_i})=d\phi_{x_j}(X_{x_j}),\;i\neq j, \end{equation*} then we can simply define the pushforward of $X$ as \begin{equation*} Y_y=d\phi_{x_i}(X_{x_i}),\;x_i\in\phi^{-1}(y). \end{equation*} And this is well-defined, since by assuming all the vectors pushforward to the same thing, even though the $X_{x_i}$'s may be unequal, the $d\phi_{x_i}(X_i)$'s are independent of the chosen $x_i$.

Also, I believe the wiki article says "A more general situation arises when $\psi$ is surjective," in reference to the case when $\psi$ is a diffeomorphism. $\psi$-relatedness is a definition.

On the other hand, we want to construct $\psi$-related vector fields. We can have the strong condition that $\psi$ is a diffeomorphism. Or, we can have this more general condition that $\psi$ is surjective, and satisfies this "projectability" condition.