Well Veeky Forums?

Well Veeky Forums?

Yoneda's lemma

Could you clarify?

p-please respond

It was a joke - that anything in category theory boils down to Yoneda lemma.

this is not category theory, it's algebraic topology

Well, I wish I could help.

I'm in PDEs/analysis.

What you want to do is prove that if [math]f \in Cov(X/Z)[/math], the [math]p_{x,y} \circ f \p_{x,y}^{-1}[/math] makes sense.
That is to say, if [math]x_1,\ x_2 \in p_{x,y}^{-1} (y_0)[/math] for some [math]y_0 \in Y[/math], then [math]p_{x,y} \circ f(x_1) = p_{x,y} \circ f(x_2)[/math].

If you do that, then [math]f \mapsto p_{x,y} \circ f \p_{x,y}^{-1}[/math] will be a morphism from [math]Cov(X/Z)[/math] to [math]Cov(Y/Z)[/math] with kernel [math]Cov(X/Y)[/math] which is exactly what you want.

Those [math]f \p_{x,y}^{-1}[/math] should be [math]f \circ p_{x,y}^{-1}[/math]

I proved this exact result last quarter
can't remember for the life of me and am too lazy to help you out

though i do remember there's a proof of this online somewhere

Yes, thank you the observation. I have been trying just this but am not succeeding. Any tips?

Any ideas on where this material could be?

This might be somewhere in the first chapter of hatcher, but I dont want to do your homework for you

I've looked around it for a while, couldn't find it. I'll keep looking.

I'm just looking for a hint, I have a general idea of the proof like noted, which is pretty clear just from basic definitions. I just can't make it work.

here

Yeah, actually I've given a little bit more thought today and it seems my idea doesn't work, since what I suggested to prove seems wrong.
Consider for instance that Z is a circle, Y is two circles and X is 4 circles.
Clearly, [math]Cov(X/Z) = \mathfrak{S}_4[/math] since you can just mix the different circles as you want, and there's no reason why two points that are in the same [math]p_x,y^{-1}(y_0)[/math] will remain in the same [math]p_x,y^{-1}(\tilde{y}_0)[/math]

Use Yoneda's lemma.

you may assume they are all path connected

Okay, then it's probably going to be true and not all that hard to prove.

snake lemma

By the way, is OP's result true if the covering spaces aren't connected ?
Is it significantly harder to prove ?

Is it then? Help pls