# Push-forward ?

1. Jun 6, 2009

### .....

"push-forward" ?

Can someone help me understand what this is, in as simple terms as possible?

If I have a function $$f: A\rightarrow B$$ and another one $$g:B\rightarrow C$$ I know the "pullback" $$f^{*}g: A\rightarrow C$$ is $$f^{*}g = g\circ f$$ (correct?)

But what about the push forward $$f_{*}g$$? What is that?

Thanks for any help.

2. Jun 6, 2009

### Hurkyl

Staff Emeritus
Re: "push-forward" ?

What is the context?

Anyways, I imagine you're just talking about functions acting on functions, in which case

f*(g) = g o f = g*(f)

3. Jun 6, 2009

### dx

Re: "push-forward" ?

Firstly, g is defined on B, so you can't push it forward with f because f: A → B. You can only push things forward with f when they are defined on A, and then too only if they are covariant. Functions are contraviarant, so even if G was a function defined on A, you cannot push it forward with f.

4. Jun 6, 2009

### .....

Re: "push-forward" ?

Thanks for your help guys...

Yes Hurkyl I'm just doing functions on functions.

dx I'm a little confused - if $$g_{*}f=g\circ f$$ as Hurkly says then we'd have f mapping A to B, then g mapping B to C ... wouldn't that be OK? Perhaps I'm missing something serious here :S ... could you give me a simple example?

Also what do you mean by covariant and contravariant?

5. Jun 7, 2009

### dx

Re: "push-forward" ?

I'm not sure what Hurkyl meant, but g o f = f*g, not gf. Contravariant objects are things that can be pulled back, and covariant objects are things that can be pushed forward. As I said, functions are contravariant, so a function on B can be pulled back by f : A → B to give you a function on A: f*g = g o f.

6. Jun 7, 2009

### Hurkyl

Staff Emeritus
Re: "push-forward" ?

Functions are contravariant on their domain, but covariant on their codomain.

Well, that's somewhat of an abuse of language. More accurately, "the set of functions from X to Y" is a functor contravariant in the variable X and covariant in the variable Y.

If I denote it by Hom(X,Y), then:

For any function f:Y->Z, I have f*om(X,Y)->Hom(X,Z) given by f*(g) = f o g

For any function f:W->X, I have f*om(X,Y)->Hom(W,Y) given by f*(g) = g o f

For an example in the setting of manifolds, recall that for a manifold M, we define a "curve on M" to be a continuous function [0,1]->M. This is a case where we fix the domain and vary the codomain, so curves get pushed around covariantly: given any continuous function f:M->N and curve c on M, we have a pushforward curve f*(c) on N, given by f*(c) = f o c.