Lemme check this: Under ZFC, regardless of CH, there is a sub-field of the reals with cardinality $\aleph_1$, right?

If CH holds, $\beth_1 = \aleph_1$, so the reals themselves are an example. If ￢CH holds, using Axiom of Choice, we choose $\aleph_1$ elements from the $\beth_1$ elements that extend $\mathbb{Q}$.

@DannyuNDos sure. Choose aleph_1 distinct elements of R and the field generated by them has size aleph_1

I've just noticed something peculiar about group completion. For consider the ordinal $\omega^2$ as an additive monoid. Since $1 + \omega = \omega$, the group completion will cancel the $\omega$ out to yield $1 = 0$.

A following conclusion is that, group completion is not necessarily injective.

yeah, you need some kind of cancellation condition to hold in the original thing for there to be any embedding in a group

otherwise, the collapsing can be very dramatic

place is very quiet without Ted the troublemaker

I'm playing with Cat.

Equalizers and coequalizers, pushouts and pullbacks, short exact sequences, and such things under Cat.

In particular, I'm testing whether the commutative diagram of forgetful functors between Ab, Grp, CMon, and Mon is a pushout square, or a pullback square, or both.

@DannyuNDos I think it's a pullback square. It should boil down to the fact that a monoid is an abelian group iff it's commutative and a group

That was my thought as well, but a proof should explicitly give $g : \mathbf{C} \to \mathbf{Ab}$ from $f_1 : \mathbf{C} \to \mathbf{Grp}$ and $f_2 : \mathbf{C} \to \mathbf{CMon}$.

Can someone explain if this (stackoverflow.com/a/64974659/2364796) makes any sense at all Linear Algebraically? I don't get "The literal position of that vector maybe garbage because by checking it in the euclidean space, you will anchor it on the origin." I get vector addition geometry, but surely the only way Queen - Woman ≈ King - Man and Queen - King ≈ Woman - Man is if the vector Queen - Women is near Royal or some similar words and Woman - Man is near some gender toggling words?

@LukasHeger I think I found the answer is negative. For let $\mathbf{C}$ be the category of commutative rings, let $f_1$ be to forget multiplicative structures, and let $f_2$ be to forget additive structures. There is certainly no $g$.