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In my differential geometry lecture today we learnt about the push-forward of a vector field by a diffeomorphism. I know some basic category theory and I noticed a functor popping up. Here's what I've got (in class at the moment we're only looking at open subsets of $\mathbb{R}^n$ but I think this ought to generalise):

Let $U,V\subseteq \mathbb{R}^n$ be open, $F\in \text{Diff} (U,V)$ a diffeomorphism, and let $\mathbb{X}:U\rightarrow \mathbb{R}^n$ be a vector field on $U$ (which we have been thinking of just as a function to $\mathbb{R}^n$). Denote the set of all vector fields on $U$ by $\mathcal{X} (U)$. Then the push-forward of $\mathbb{X}$ by $F$ is the vector field $F_{*} \mathbb{X}\in\mathcal{X} (V)$ defined by $(F_{*} \mathbb{X})\circ F = F'\mathbb{X}$ where $F'$ is the derivative of $F$. In components, $(F_{*} \mathbb{X})^i (y) = \partial_j (F^i (F^{-1} (y)))\mathbb{X}^j (F^{-1} (y))$. You can check that ${id_U}_{*} = id_{\mathcal{X} (U)}$ and if $U\xrightarrow{F} V\xrightarrow{G} W$ is a sequence of diffeomorphisms then $(G\circ F)_{*}\mathbb{X} = G_{*} F_{*} \mathbb{X}$.

Therefore we've got a covariant functor $h_{*}(F:U\rightarrow V) = F_{*} : \mathcal{X}(U)\rightarrow \mathcal{X}(V)$ from the category $\bf{Man}$ of smooth manifolds (here I just used open subsets of $\mathbb{R}^n$) to the category $\bf{VecFld}$ of vector fields on smooth manifolds. However, I don't have a clue what $\bf{VecFld}$ looks like as a category itself. What are the morphisms? Is every morphism $f:\mathcal{X}(V)\rightarrow \mathcal{X}(U)$ the pullback $f(\mathbb{X}) = \mathbb{X} \circ F$ of some diffeomorphism $F:U\rightarrow V$? Does $h_{*}$ have an adjoint?

Any information on related questions would be greatly appreciated, and sorry if this is very basic; I hope to learn more about vector fields later in this course!

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Here are at least two good ways of viewing the pushforward as a functor.

First, consider the category of pointed smooth manifolds, where the objects $(M,p)$ are smooth manifolds $M$ with a distinguished point $p$, and the morphisms $F:(M,p)\to(N,p')$ are smooth maps preserving the distinguished points, i.e. such that $F(p)=p'$. Then the pushforward of $F$, $F_*$ is a linear map between tangent spaces, $F_*:T_p M\to T_{p'} N$. So we can think of the pushforward as a functor from pointed smooth manifolds to the category of vector spaces with linear maps.

Second, consider the category of smooth manifolds with diffeomorphisms as morphisms. Let $F:M\to N$ be a diffeomorphism. We have the pushforward, as defined above, for each point of $p\in M$, $F_*:T_p M\to T_{F(p)}N$. Letting $p\in M$ vary this gives a bundle map between tangent bundles, $F_*:TM\to TN$. So here we can think of the pushforward as a functor from the category of smooth manifolds with diffeomorphisms to the category of smooth vector bundles with smooth bundle maps.

Note in the last example we need to take diffeomorphisms as morphisms because otherwise we don't get a well defined bundle map. This relates to why we can't pushforward a vector field with a map which isn't a diffeomorphism.

I realize these examples may be a bit too advanced for you right now but I believe they are the most basic ways to view the pushforward as a functor. In time you will learn about all the concepts I mentioned above.

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  • $\begingroup$ You didn't say a thing about vector fields which is what the question is about. $\endgroup$ – Omar Antolín-Camarena Oct 16 '14 at 18:01
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One way to say in categorical terms what your remarks show is that there is a functor from the category $\bf{Man}^\simeq$ of smooth manifolds, where the morphisms are diffeomorphisms, to the category of sets; this functor assigns to a manifold the set of vector fields on it, and assigns to a diffeomorphism $F$ the pushforward function along $F$.

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  • $\begingroup$ Oh yes - I see that in my "functor" I've gone one level too far down and forgotten that it maps each manifold to the set of vector fields on it! Many thanks $\endgroup$ – Alex Saad Oct 16 '14 at 18:49

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