The current problem set in algebraic geometry is making me think about something you'd probably approve of instead of all the algebra: the Veronese embeddings
if you want a compact definition for modules, you don't need the whole category stuff, just define them as an abelian group $A$ together with a ring homomorphism $R \to \mathrm{End}_{\Bbb Z}(A)$
Jokes apart I'm kinda getting used to the categorical language between algebraic geometry and algebraic topology. I still have to wrap my head around adjoint stuff at some point though
Suppose you can launch a projectile with speed $v_0$ at any angle towards a wall, which is a horizontal distance $L$ away. What's the highest point on the wall you can hit?
speaking of categorial nonsense - the centre of a ring $A$ is isomorphic to the endomorphism ring of the identity functor on left (or right? I think it doesn't matter but I'm too lazy to check) $A$-modules
for what it's worth i read that paper right around the time I read Gugenheim-May's paper (really fantastic imo and very useful for me personally) and it gave me really good intuition / understanding for both the model categorical phrasing of things and also even some of the computational aspects
@MatheinBoulomenos Thanks for the offer but I'm a bit busy right now so I wouldn't really be able to follow you. I might ask if the offer is still valid in a while though!
@MatheinBoulomenos Yeah! Anyway - recently I learnt (although maybe part of this is wrong) that Hochschild cohomology of an algebra is isomorphic to the endomorphism ring of the identity functor on the derived category, hence is the "derived" version of centre
(btw, when I say 'range formula', I mean "how far can you reach on flat ground using any angle you want', not 'how far will you reach on flat ground using a given angle')
for the latter it's best just to do the direct computation, i.e. "how long will it remain in air, and how far will it travel horizontally in the meantime"
to put it differently: the usual computation has you consider the ground as a horizontal boundary. I'm suggesting you take the wall as a vertical boundary
Start as with any kinematics problem: What are $x(t)$ and $y(t)$ as functions of time $t$? (There's also $v_x,v_y$ as functions of time but they're not terribly pertinent here.)
speaking of categorial nonsense - the centre of a ring $A$ is isomorphic to the endomorphism ring of the identity functor on left (or right? I think it doesn't matter but I'm too lazy to check) $A$-modules
endomorphisms is a hom functor evaluated with the same object in two arguments. It's logical that it's not functorial since the hom functor is covariant in one argument and contravariant in the other one
so if $C$ is any category we can consider the fulll subcategory of $C \times C^{op}$, consiting of objects of the form $(M,M)$, then taking endomorphisms in $C$ is functorial from that category, since a morphism $M \to N$ in that category is a pair of morphisms $M \to N$ and $N \to M$. there's a functor from the subcategory of $C$ where we only allow isomorphisms as functors sending $M$ to $(M,M)$ and $f$ to $(f,f^{-1})$
If $d(x,y)$ and $d'(x,y)$ are a metrics on $X$, then $d'(x,y)=\frac{d(x,y)}{1 + d(x,y)}$ and $d(x,y)$ generates the same topology.
My attempt:-
Let $\mathscr T$ be the topology generated by $d$ and $\mathscr T'$ be the topology generated by $d'$. Let $U\in \mathscr T\implies U=\bigcup_{\alpha\...
If $d(x,y)$ and $d'(x,y)$ are a metrics on $X$, then $d'(x,y)=\frac{d(x,y)}{1 + d(x,y)}$ and $d(x,y)$ generates the same topology.
My attempt:-
Let $\mathscr T$ be the topology generated by $d$ and $\mathscr T'$ be the topology generated by $d'$. Let $U\in \mathscr T\implies U=\bigcup_{\alpha\...
