@Perturbative a ring homomorphism $R \to S$ induces a faithful functor $\textbf{$S$-Mod} \to \textbf{$R$-Mod}$ that has a left adjoint...

the ring homomorphism perspective of module just makes this easier

@ÍgjøgnumMeg another perspective is that inclusion is continuous in the subspace topology by definition

anyone want to explain in everyday words the relationship between group (actions?) and symmetry?

music-theory (See above.)

@Shaun I don't understand the joke.

@JohnNash That's okay, I'm kind of lame :)

When you spent a lot of time on an easy answer but don't get any upvotes . . .

@mathwannabe Try asking in the room below.

@LeakyNun hi

hi

@AkivaWeinberger old

hi chat

$$\newcommand{sinc}{\operatorname{sinc}}\sum_{n=1}^\infty \sinc n = \sum_{n=1}^\infty \sinc^2 n$$

$$\int_0^\infty \sinc x \ \mathrm dx = \int_0^\infty \sinc^2 x \ \mathrm dx$$

@AkivaWeinberger Recently the drones at Gatwick airport left people stranded for days because the planes could not fly.

@LeakyNun I guess the more interesting question is: What is the mechanism that enforce this equivalence

also conjecture:

$$\int_0^\infty \sinc x \ \mathrm dx = \int_0^\infty \sinc^{2n} x \ \mathrm dx$$

Another conjecture:

$$\lim_{n\to \infty} \int_0^{\infty} \text{sinc}^{2n} x dx = \int_{0}^{\infty} \delta (x) dx = 1$$

@AkivaWeinberger XD

@Shaun thanks for the pointer

Hi

@mathwannabe You're welcome :)