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.
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 $g$.
@user10478 I am not sure but it feels like they are saying that the 4 points( queen, king, woman, man) are the vertices of a parallelogram. So the opposite sides represent the same vector. And I guess by anchoring it on the origin, they mean affine transformation of the plane to slide one of the vertices to origin.
