Thread: Push me, Pull you

So, in another thread I mentioned the pull-back in passing. Let's do it properly now, as it is just so much fun.

We will suppose that are vector spaces, and that the linear operators (aka transformations) . Then we know that the composition as shown here.



Notice the rudimentary (but critical) fact, that this only makes sense because the codomain of is the domain of

Now, it is a classical result from operator theory that the set of all operators is a vector space (you can take my word for it, or try to argue it for yourself).

Let's call the vector space of all such operators etc. Then I will have that are vectors in these spaces.

The question naturally arises: what are the linear operators that act on these spaces? Specifically, what is the operator that maps onto ?

By noticing that here the "" is a fixed domain, and that , we may suggest the notation . But, for reasons which I hope to make clear, I will use a perfectly standard alternative notation .

Now, looking up at my diagram, I can think of this as "pushing" the tip of the "f-arrow" along the "g-arrow" to become the "composite arrow". Accordingly, I will call this the push-forward of on , or, by a horrid abuse of English as we normally understand it, the push-forward of

So, no real shocks here, right? Ah, just wait, the fun is yet to begin, but this post is already over-long, so I'll leave you to digest this for a while.........

Wonderful stuff! I look forward to the beginning of the fun.

Since it seems like the topology thread is wrapping up, you can consider me on board for this thread for the time being.

What do you mean “wrapping up”? What’s wrong with it?

Originally Posted by JaneBennet

What do you mean “wrapping up”? What’s wrong with it?

Originally Posted by Guitarist

This is about as far as I want to go with this rather sketchy romp over topology. You at least have enough grounding to make some sense of anything you choose to read up on the subject. Beyond this point it gets rather specialized, and, to me, not too interesting.

We also have enough grounding to make sense of the theory of manifolds, which I am tempted to start in an equally superficial way.

Just going by the above. ^

Ah, okay. :-D I was worried that something was wrong.

Recall that, given as a linear operator on vector spaces, we found as the linear operator that maps onto , and called it the push-forward of .

In fact let's make that a definition: defines the push-forward.

This construction arose because we were treating the space as a fixed domain. We are, of course, free to treat as as fixed codomain, like this.



This seems to make sense, certainly domains and codomains come into register correctly, and we easily see that .

Using our earlier result, we might try to write the operator , but something looks wrong; is going "backwards"!

Nothing daunted, let's adopt the convention . (We will see this choice is no accident)

Looking up at my diagram, I can picture this a pulling the "tail" of the "h-arrow" back along the "g-arrow" onto the composite arrow, and accordingly (using the same linguistic laxity as before), call the pull-back of , and make the definition: defines the pullback (Compare with the pushforward)

This looks weird, right? But it all makes beautiful sense when we consider the following special case of the above.



where I have assumed that is the base field for the vector spaces. As before, the composition makes sense, and I now have , and the pullback . But, hey, lookee here....

is the vector space of all linear maps , so we quite simply have that , the dual vector spaces.

Putting this all together I find that, for I will have as my pullback.

I say this is just about as nice as it possibly could be.

I have one further trick up my sleeve.........

Wow I really wanna learn some of this, but I'm going on holiday for two weeks on saturday, and I want to make sure I'm understanding all the set stuff nice and soundly before I jump into topology and this. I'm not gonna be active in these threads (or at all) over the next two weeks, but will jump in after that. Anywya, thanks for making them.

Being such a sweet guy, I am taking Mary-Bernadette out for dinner tonight. So, while she is "getting ready" (how long need that take?), I can't resist delivering my punch-line.

Recall that we had some gadget , the map from all vector spaces to their duals, and, for all , we had . the map from all linear operators on to their pullbacks.

From the immediately foregoing, we may easily see that, if there is some identity map , then as the identity map

This is quite simply the definition of a functor, that is is the functor from the the "category" of vector spaces to the "category" of their duals.

I think that is soooo cool, or, as we like to say over here- it's the dog's dangleys!