Let $R$ be a noetherian ring and $I \subseteq R$ an ideal, and $S \subset (R\setminus \{ 0\}, 1, \dot)$ a submonoid with $S \cap I = \varnothing$. Then $0 \to I \hookrightarrow R \to R/I \to 0$ is a short exact sequence of f.g. $R$-modules, so $\operatorname{Hom}_{S^{-1}R}(S^{-1}(-), S^{-1}N)$ is exact and we get that $\operatorname{Hom}_{S^{-1}R}(S^{-1}I, S^{-1}N) \subseteq \operatorname{Hom}_{S^{-1}R}(S^{-1}R, S^{-1}N)$..
And then Baer it
So $S^{-1}N$ is injective in $_{S^{-1}R}\text{Mod}$ ?
I'm trying to arrange tiles with a total area of 125 units into a 2d shape. Since 125 is not a square or rectangular number, they would not fit there. Any suggestions?
if you take a number of independent random variables with the same distribution, add them up and properly normalize this, you will get something that's Gaussian in the limit
Unlike in English, there is a thing called gender, and in German and Latin there are three options, not just two. das is neuter, der is masculine, die is feminine.
No, no. $b^3/3$ is the UNIQUE number $A$ satisfying $s_n<A<S_n$ for all $n$.
Since $S_n-s_n$ can be made arbitrarily small by making $n$ big enough, there can only be one single number $A$ that satisfies that inequality for all $n$.
How do connected sums work? Say we have two manifolds $M,M^{\prime}$ and charts $\varphi\colon B^n\rightarrow M$, $\varphi^{\prime}\colon B^n\rightarrow M^{\prime}$. Then we glue together $M\setminus\varphi(0)$ and $M^{\prime}\setminus\varphi^{\prime}(0)$ by identifying $\varphi(x)\sim\varphi^{\prime}((1-||x||)x)$. On its own, I think this is just identifying one ball minus origin with the other ball minus origin turned inside out, but what does this actually do when these balls are embedded into manifolds? And why do we need to do it this way?
Like, supposing he said $\frac{b^3}{4}$ for the area, could we still somehow arrive at $\frac{b^3}{3}$ being the one true area (given that it is applicable for all $n \ge 1$)?
No, you discover the $b^3/3$ from these formulas and then you check, as I said, that there's a unique number that works because $S_n-s_n$ gets arbitrarily small.
@Naganite There is one and only one number that is below all upper sums and above all lower sums (in the event that you have something where it makes sense to integrate). If it depended on n, there'd be infinitely many such numbers, and therefore there wouldn't be a well-defined area.
And I mean maybe this is opening a can of worms on its own, but then doesn't that beg the question as to whether or not we can assume there is only one such area?
Yes. That's why the funny stuff happens with the gluing.
In algebraic geometry, this shows up explicitly when you talk about blow-up of a point in a complex $n$-manifold and this turns out to be connect sum with a $\Bbb CP^n$ with reverse orientation.
@Naganite I see that Apostol has not yet defined upper and lower sum rigorously. No matter---divide the interval [0,1] into pieces. The upper sum for that division is where you make the highest rectangles possible on those pieces, and the lower sum is the one where you make the lowest rectangles possible on those pieces.
Hey guys, would love to get some direction on the following problem I have to solve. The concept is that a Monster is positioned at (x1, y1) and the player is positioned around him (x2, y2). Once the Monster attacks the player, the player must get pushed backward by 3 points.
So given a slope - (y1 - y2) / (x1 - x2), starting point = (x2, y2) and center point (x1, y1), and d is the distance that the new point should be at.
So will doing the gluing "naively" as in my picture give something diffeomorphic to the thing we get from doing the gluing properly, but simply not carry a natural orientation consistent with the orientation of the summands?
in this specific case, it's clearly still a sphere (as it should be)
@Naganite To be more precise: if you've split the interval into pieces, the upper sum is where each rectangle's height is the supremum of the function in each piece, and the lower sum is where each rectangle's height is the infimum et cetera.
Yeah, sorry, it was too late to edit the example. Ted pointed this out too.
if you do the gluing in my above picture, you still get a sphere, which is orientable, but either orientation on the sphere will not extend the orientation of one of the summands in that picture
Let $f: X \to \{x_0\}$ and $g : Y \to \{y_0\}$ be homotopy equivalences. Define $F : X\times Y \times I \to \{x_0\} \times \{y_0\} \times \{0\}$, $F(x, y, s) = (f(x), g(y), 0)$. This descends to the quotient $X\ast Y$ and has an obvious homotopy inverse.
I never said deformation retract, @feynhat, did I? Still, the standard definition is that the identity map is homotopic to a constant map. Isn't that what "homotopy type of a point" means?
