For nearly 100 y from the advent of General Relativity (GR) in 19015, no book author could advance accurately GR with all mathematical details. This is the reason, a large number of anxious reader still don’t understand GR to the appreciating level. Most physicists are badly unskilled in mathematics. In order to hide their incompetency they become more and more wordy. While a single equation can be more weighty than a dozen page of wordy writing. ]]>

By the way, I really appreciated the additional remark about the intuition behind the theorem. It shows the cleverness of the theorem, which is indeed quite beautiful actually. ]]>

I think you’re actually running right up onto the deep meaning of the existence theorem, even if you don’t quite realize it yet. The whole idea is that being a cone is actually *equivalent* to equalizing and . That is, these two arrows encode two different operations, and the point of a cone is that the two are equal.

Remember that a cone consists of an object and a family of arrows , one for each object . But not any such collection works; there’s a naturality condition! For we have .

So, the real trick of the existence theorem (which maybe I could have explained better before), is that if you’ve got products and equalizers, you can encode this naturality condition in terms of the equalizer of two arrows in ! A cone, then, is exactly a collection of arrows such that the unique arrow into the product equalizes the two arrows and . And a limit — a universal cone — is just the equalizer of the two arrows.

The key, then, is to understand how the arrows and encode the naturality condition. I’ve actually done this once above, in the particular case of the equalizer, when I checked that it was a cone itself. What’s left is just to run that same argument *backwards*, showing that any cone equalizes the two arrows.

So, re-draw the left-hand triangle of the diagram, but instead of , use . The vertical arrow is , and it should be clear why . Being a cone tells us that for every

which we rewrite as

Exactly as before, we can see that and that . So again we rewrite:

Now, since that’s true for all , the product property (over the morphisms of ) tells us that

Just as we wanted!

]]>Could I ask you for some clarifications about the very last passage?

How do you check than $fh = gh$? I understand that then we get another cone of the same type of the equalizer of $f$ and $g$, and thus we find the factorization we needed, but to do that we still have to guarantee that commutativity condition.

I thought about decomposing $h$ in the $\lambda_J$s that generates it and checking the composition “component-wise”, but I don’t know how to do this check. I suspect that the equation $\lambda_J = F(u) \circ \lambda_K$ might be involved although I didn’t manage to plug it in the correct way to get the result. ]]>

Or.. don’t, if these notes are insufficient to your lofty standards…

]]>How about these tables:

]]>Actually the fact is that if it did NOT, mathematicians would not be interested in THIS theory–so not “lucky” but necessary

]]>In a way, you can think of them in programming terms as casting up an inheritance hierarchy: yes, is a monoid, but if all we care about is its set of elements we can just consider .

The really interesting thing is that forgetful functors are usually one side of an adjunction, and the other side is a “free” functor, that gives the “most generic” way of building a new layer of structure on top of a given one. If is a set, then the free monoid on is the most generic monoid that includes the elements of .

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