Is there a relationship between the pullback in differential geometry and that in category theory?

Question

1. Is there a relationship between the pullback in differential geometry and the pullback in category theory?

[2. Is there a relationship between the pushforward/pushout in differential geometry and the pushforward/pushout in category theory? Although the answer to the above (1.) is equivalent to the answer to this question (2.) by duality.]

As far as I can tell, (for the terms in differential geometry) the pullback is a contravariant functor, and the pushforward is a covariant functor.

3. Is there a way to turn every contravariant functor into a category theory pullback or vice versa? (Or every covariant functor into a category theory pushforward or vice versa?)

If the answer to 3. is no, then it seems like the fact that the two concepts have the same name is just a historical accident, and does not indicate that one is meant to generalize the other.

4. Is it true that the similar terminology between the two fields is a historical accident? Or are they both meant to evoke the same type of basic example?

Answer 1

"Pullback" and "pushforward" are often used when you have a map $f:X \to Y$ and do things with structures built on $X$ and $Y$.

The pullback of category theory can be viewed as the specific case where you're pulling "maps with codomain $Y$ back to "maps with codomain $X$". In fact, in a Cartesian category $\mathcal{C}$, the pullback can be used to define a functor $f^* : \mathcal{C}/Y \to \mathcal{C}/X$.

One particular example we might consider is in Top: if we have a bundle $E \to Y$, then the pullback bundle $f^*E$ is precisely the category theoretic pullback. I have some belief that this example is the actual etymology of the term, and many actual uses of pullbacks can be viewed as having the same flavor.

Also, I think the case where you have two bundles $E_1 \to Y$ and $E_2 \to Y$ and form the bundle $E_1 \times_Y E_2 \to Y$ is the reason the pullback is sometimes called the "fiber product".

Answer 2

Consider the diagram in the arrow category of smooth manifolds: $$\begin{array}{rcccl} & & \varphi^* & & \\ & \swarrow & & \searrow & \\ \pi_N & & & & \pi_M \\ & \searrow & & \swarrow & \\ & & \varphi & & \end{array} $$ Provided one can show that $\pi_N=\varphi\circ\pi_M\circ\varphi^*$, one has that there is a left arrow in this diagram, the right arrow is simpler, then the diagram commutes trivially in the arrow category.

Left arrow: $$ \begin{array}{rcccl} \, & T^*N & \overset{\varphi^*}{\rightarrow} & T^*M & \, \\ id_{T^*N} & \downarrow & & \downarrow & \varphi\circ\pi_M \\ \, & T^*N & \overset{\pi_N}{\to} & N & \, \\ \pi_M\circ\varphi^* & \downarrow & \, & \downarrow & id_N \\ \, & M & \overset{\varphi}{\to} & N & \, \end{array}$$

Right arrow: $$ \begin{array}{rcccl} \, & T^*N & \overset{\varphi^*}{\to} & T^*M & \, \\ \varphi^* & \downarrow & \, & \downarrow & \pi_M \\ \, & T^*M & \overset{\pi_M}{\to} & M & \, \\ \pi_M & \downarrow & \, & \downarrow & \varphi \\ \, & M & \overset{\varphi}{\to} & N & \, \end{array}$$

Having shown all this, we just would need to verify that the universal property of the category theoretic pullback is satisfied. Then it would follow that the pullback in differential geometry, in the arrow category of smooth manifolds, is a category theoretic pullback.