After all, one of the major reasons diff forms is useful is because pullback puts the whole change of variables theorem very neatly, or so I understand.
First we shall do it for the restricted case then as you said we can easily generalize. In the maps below I define it on basis of the group and extend the map. Consider $\phi : Z_4 \rightarrow Z_8 \oplus Z_2$ where $1 \mapsto (2,1)$ the $Im(\phi) = <(2,1)>$
The map above is clearly injective. I will show now that $(Z_8 \oplus Z_2) / Im(\phi)$ is cyclic group of order 4.
It is clear that it has order 4.
The power of the coset (1,1) + H is : (1,1) + H, (2,0) + H, (3,1) + H, (4,0) + H = H. Hence this coset is of order 4 so $Z_8 \oplus Z_2 / Im(\phi) = <(1,1) + H>$
@balarka following our earlier discussion, am I right in thinking that the $dt$ in that question (i.e. in $\int \omega_{\gamma}\gamma'(t)\,dt$) should not be interpreted as a 1-form?
So the map $\theta : Z_8 \oplus Z_2 \rightarrow Z_4$ defined as $(1,1) \mapsto 1$ is clearly surjective and that it has the property that $(Z_8 \oplus Z_2) / Im(\phi) = Z_4$
@ForeverMozart First I was worried because in general two cardinals aren't comparable without choice (and we aren't assuming choice here) but can't I just use the definition of $<$ on cardinals as there existing an injection?
We pulled back a form $\omega$ by some parametrization $[0, 1] \to \gamma$. That gives me $\text{something} dt$. That's a 1-form on $[0, 1]$, for sure.
@Semiclassical The reason people call that a 1-form is not because forms give a new way of integrating (indeed, when one says "integrate the 1-form $dx$" it literally just means take $\int dx$ in the standard way". It's because it transforms via change of variables the same way a 1-form does under pullback.
The way "it eats vectors!" shows up is that vectors don't transform in quite the same way under coordinate change that integrals do (the chain rule). But their duals do.
it just feels like there's a shifty split somewhere in there, where certain expressions get interpreted as 1-forms and others as symbols defining a Riemann sum
@BalarkaSen If I have the following map $(Z_8 \oplus Z_2) / Im(\phi) \rightarrow Z_4$ defined by $(1,1) + Im(\phi) \mapsto 1$ this map is clearly isomorphism.
How can I use that map to define a surjective map between $(Z_8 \oplus Z_2)$ and $Z_4$
@Adeek You should probably think through this, because I am not really paying too much attention to what you are saying :) As I said, maybe ask someone else who'd be able to give a thorough walkthrough.
What MikeM was saying, I believe, is if you can see how that integral behaves under u-subs - that is, if you have that information, then you can tell $dx$ is a 1-form.
@SemiC E.g., $ds$ doesn't behave well under u-subs, does it?
@BalarkaSen let us say I have $\theta : A \rightarrow B$ we know by first isomorphism theorem if $\theta$ is surjective then $A/B ~= B$ but can you go the other way suppose you have a map $f : A/B \rightarrow B$ that is an isomorphism can you get $\theta$ from that ?
@balarka well, i think the point is that $ds$ is better thought of as a notational convenience, i.e. that it's simpler to write that than $\sqrt{dx^2+dy^2}$
@Anthony I suspect that you will run into problems. If large cardinals exist, aren't they counterexamples? Since their existence is independent of ZF, it really does not matter if they are assumed to exist or not: any proof you want wouldn't be possible.
