Good introductory books on homological algebra
Good textbooks on homological algebra
Correct use of 'with' or 'at' when talking about percentages? I'm writing the following sentence:     Women lead uptake at XX%.   But a friend is saying that it should be written:     Women lead uptake with XX%.   Which one is correct? I did a dumb thing and now need to undelete some files.  In fact, I have already recovered the files using , which has served me well on the other rare occasions where this has been necessary.  So I have the files.    The problem is that Glary and the  don't preserve the original folder structure.  In this case I have about 3,000 small files that I must match to their original locations in a complex nested folder hierarchy.  I could see the full folder paths in Glary before the restore, so I know the information is there; it's just not using that information and copying everything to root of my designated destination, rather than relative to the original location. Altium: How can I edit a component from a library When viewing a question, when I hover over the star icon (under the voting module) the tooltip says I've made it a favorite when I haven't yet. Why does the natural ring homomorphism induce a surjective group homomorphism of units? I'm trying an old problem here:   Suppose $n\mid m$, and I have a natural ring homomorphism $\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $\varphi(j)=j\mod{n}$. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups $(\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times$?  My question in particular is why is it surjective? I take $k\in(\mathbb{Z}/n\mathbb{Z})^\times$. Then there exist integers $s,t$ such that $sn+tk=1$. Since $n\mid m$, I can also write $m=na$. So multiplying through I get $sm+tka=a$. I'm lost here. How can I find a unit in $(\mathbb{Z}/m\mathbb{Z})^\times$ which maps to $k$ to show surjectivity? Alternative IDE for VB6 and VBA I've been spoiled by Visual studio 2008 and Eclipse and have to do a little maintainence work on a VB6 app.  Does anyone know of an alternative/ updated IDE for VB6?  A rewrite is not an option I'm just fixing a couple of bugs and it's a big codebase. Show that AB is singular if A is singular Actually I need to show that $\det(AB) = \det(A)\det(B)$ if $A$ is a singular matrix.  The determinant of $A$ is $0$ if $A$ is singular, so $\det(AB)$ has to be $0$ as well, but I have problems showing that $AB$ is singular if $A$ is singular. How can I show that? Can not move polygons but only vertices of an object by bmesh! Show that $\cot \frac{\pi}{2m}\cot \frac{2\pi}{2m}\cot \frac{3\pi}{2m}...\cot \frac{(m-1)\pi}{2m}=1$ What is the difference between public, protected, package-private and private in Java? Why is the Earth so fat? I made a naive calculation of the height of Earth's equatorial bulge and found that it should be about 10km.  The true height is about 20km.  My question is: why is there this discrepancy?  The calculation I did was to imagine placing a ball down on the spinning Earth.  Wherever I place it, it shouldn't move.  The gravitational potential per unit mass of the ball is $hg$, with $h$ the height above the pole-to-center distance of the Earth (call that $R$) and $g$ gravitational acceleration.  Gravity wants to pull the ball towards the poles, away from the bulge.  It is balanced by the centrifugal force, which has a potential $-\omega^2R^2\sin^2\theta/2$ per unit mass, with $\omega$ Earth's angular velocity and $\theta$ the angle from the north pole.  This comes taking what in an inertial frame would be the ball's kinetic energy and making it a potential in the accelerating frame.  If the ball doesn't move, this potential must be constant, so  $$U = hg - \frac{(\omega R \sin\theta)^2}{2} = \textrm{const}$$  we might as well let the constant be zero and write  $$h = \frac{(\omega R \sin\theta)^2}{2g}$$  For the Earth,   $$R = 6.4*10^6m$$  $$\omega = \frac{2\pi}{24\ \textrm{hours}}$$  $$g = 9.8\ m/s^2$$  This gives 10.8 km when $\theta = \pi/2$, so the equatorial bulge should be roughly that large.    , the Earth is 42.72 km wider in diameter at the equator than pole-to-pole, meaning the bulge is about twice as large as I expected.  (Wikipedia cites diameter; I estimated radius.)  Where is the extra bulge coming from?  My simple calculation uses $g$ and $R$ as constants, but neither varies more than a percent or so.  It's true the Earth does not have uniform density, but it's not clear to me how this should affect the calculation, so long as the density distribution is still spherically-symmetric (or nearly so).  (Wikipedia also includes , without derivation, that agrees with mine.) How to shut down the computer safely by pressing the power button? So this is pretty simple: when I press the power button whilst the machine is on, it comes up with a menu (of which one of the options is to shutdown)  However, I just want it to just start shutting down once I press the power button without any further interaction!  How do I accomplish this? On September 29th, I created the  tag for  Today, while checking the tag description, I saw that the  seems to be .  Why is that? Why has the Community user "stolen" the spot as author of the tag? Validation Error less than training error? I found two questions  and  about this issue but there is no obvious answer or explanation yet.I enforce the same problem where the validation error is less than training error in my Convolution Neural Network. What does that mean? How to find max of amount of opportunity related to account? I have a below requirement : I need to find maximum amount on related opportunities and to display on account. Ex: suppose I have account A1 and A1 have 5 opportunities with amount o1=>100 o2=>500 o3=>300 o4=>1000 o5=>50 on my account i should have max_opportunity_amount =1000.  I have written below trigger ,but  I am not sure what to do further  trigger maxOppAmountonAcc on Opportunity(after insert){      map&lt;id,Opportunity&gt; oppMap = new map&lt;id,Opportunity&gt; ();     set&lt;Id&gt; oppId = new set&lt;Id&gt;();     for(Opportunity o : Trigger.new)     {         oppMap.put(o.accountId,o);         oppId.add(o.id);     }       List&lt;Account&gt; accList = new List&lt;Account&gt;();      for(Account a : [Select id ,MaxOppAmount__c,(select id,max(amount)  from opportunities) from Account where id in : oppMap.keyset()])     {      }  } What points in the complex plane satisfy $|z|=arg(z)$. How to edit command line in full screen editor in ZSH? Let $a=p_1^{x_1}p_2^{x_2}\cdots\cdot p_q^{x_q}$ and $b=p_1^{y_1}p_2^{y_2}\cdots p_r^{x_r}$ ,   say — Since gcd(a,b) is the largest common divisor of a and b and is divisible by any other common divisor of the two,   GCD$(a,b)=p_1^{\min(x_1,y_1)}p_2^{\min (x_2,y_2)} \cdots p_q^{\min (x_q,y_q)}$  LCM$(a,b)= p_1^{\max(x_1,y_1)}p_2^{\max(x_2,y_2)}\cdots p_r^{\max (x_r,y_r)}$  I understand GCD(a,b) has to divide both $a,b$. Therefore the exponent of any prime factor $p_i$ in GCD has to be in both $a,b$  — therefore $p_i^{\min(x_i, y_i)}$. But I'm muddled and anxious. Why does $\min$ appear in GREATEST common divisor? Why does $\max$ appear in LOWEST common multiple? Show $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ for finite $A$ I have a finite abelian group $A$ and I have shown that $A\otimes\mathbb{Q} \cong \{0\}$, the trivial group.  Now I want to show that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$, i.e we only have the trivial zero group homomorphism from $A$ to $\mathbb{Q}$.  The question I'm trying to solve is phrased in such a way that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$ should fall out almost automagically once you have that the tensor product collapses but I'm really stuck on this even though some people I talked to said it was really easy.  Some thoughts:   I've tried to assume the existence of a nontrivial homorphism $A \to    \mathbb{Q}$ to generate a contradiction but with no success. We have the Tensor-hom adjunction theorem stating that for abelian groups $A,B,C$ we have a natural bijection  $$\text{Hom}_{\textbf{Ab}}(A\otimes B,C) \cong \text{Hom}_{\textbf{Ab}}(B ,\text{Hom}_{Ab}(A,C))$$ and by plugging in our particular groups we get that:   $$\text{Hom}_{\textbf{Ab}}(A\otimes \mathbb{Q},\mathbb{Q})\cong\text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q}))$$  Since $A\otimes\mathbb{Q} \cong \{0\} \implies \text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q})) = \{0\}$ so we only have the trivial morphism from $\mathbb{Q} \to \text{Hom}_{Ab}(A,\mathbb{Q})$ but I don't think this implies that $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ and I'm not sure if this reasoning helps me at all. I'm having a problem with author display for a statute - the author is 'City and County of San Francisco' but the output (using the apalike.bst style file provided by Elsevier) gives the author as 'City and San Francisco', missing the 'County' in the authorname.  Any suggestions?