Jabotinsky IL and Nixon's program: a first categorical foundation

[JmsNxn]

OK, I'll take time to digest this. But to be clear, I don't still understand then how can you use the omega notation on that set you define.
....


Maybe it's just an abuse of notation? When I see that Omega notation I picture it as

YES! That's absolutely it Mphlee!

Everything you are saying is correct. That's exactly how you should visualize it.

It is not an abuse of notation, for the simple reason we are MODDING OUT by the equivalence relation,

\(
\int_\gamma \simeq \int_\varphi\\
\)

The fact that this modding out works so well, is because of the following.

There is some representatives \( \gamma_1,\gamma_2 \) of \( \text{Rsd}(f,\zeta_1),\text{Rsd}(f,\zeta_2) \) respectively, such that,

\(
\int_\gamma f(s,z) = \int_{\gamma_1} \bullet \int_{\gamma_2}\\
\)

BUT, there are also some representatives \( \gamma_1^*,\gamma_2^* \) of \( \text{Rsd}(f,\zeta_2),\text{Rsd}(f,\zeta_1) \) respectively, such that,

\(
\int_\gamma f(s,z) = \int_{\gamma_1^*} \bullet \int_{\gamma_2^*}\\
\)


So when I write,

\(
\int_\gamma f(s,z)\,ds\bullet z = \Omega_{j} \text{Rsd}(f,\zeta_j;z)\bullet z\\
\)

I mean, there is SOME representative of each Rsd, where this is true. Think about modular arithmetic.

\(
x \equiv y\,\,(\text{mod} \,m)\\
\)

means there is SOME k such that,

\(
x-y = km\\
\)

It doesn't mean for all k.

So when I write \( \text{Rsd} \); I'm assuming we're in the modded out space. And when we pull back, there is SOME representative from this set where it is true. You can see this even in your fancy picture. Just reorganize the curves to put different singularities first.