Let $M^{\prime}\xrightarrow{f}M\xrightarrow{g}M^{\prime\prime}\rightarrow 0 \DeclareMathOperator{\Hom}{Hom}$ be a sequence of $A$-modules and homomorphisms. I want to show that the sequence is exact if and only if for all $A$-modules $N$, the sequence $$0\rightarrow \Hom(M^{\prime\prime},N)\xrightarrow{\overline{g}} \Hom(M,N)\xrightarrow{\overline{f}} \Hom(M^{\prime},N)$$ is exact.  I have shown that $M^{\prime}\xrightarrow{f}M\xrightarrow{g}M^{\prime\prime}\rightarrow 0$  is exact implies $0\rightarrow \Hom(M^{\prime\prime},N)\xrightarrow{\overline{g}} \Hom(M,N)\xrightarrow{\overline{f}} \Hom(M^{\prime},N)$ is exact for all $A$-modules $N$.   Now suppose for all $A$-modules $N$, $0\rightarrow \Hom(M^{\prime\prime},N)\xrightarrow{\overline{g}} \Hom(M,N)\xrightarrow{\overline{f}} \Hom(M^{\prime},N)$ is exact. I want to show that $M^{\prime}\xrightarrow{f}M\xrightarrow{g}M^{\prime\prime}\rightarrow 0$ is exact. That is I have to show $g$ is onto and $\ker(g)=\operatorname{Image}(f)$. In Atiyah Macdonald's Commutative Algebra, it is written that since $\overline{g}$ is injective for all $A$-modules $N$, therefore $g$ is onto. But I could not understand it. Any explanation will be appreciated.
The following is the first part of Proposition 2.9 in "Introduction to Commutative Algebra" by Atiyah &amp; Macdonald.     Let $A$ be a commutative ring with $1$. Let $$M' \overset{u}{\longrightarrow}M\overset{v}{\longrightarrow}M''\longrightarrow  0\tag{1} $$ be sequence of $A$-modules and homomorphisms. Then the sequence (1) is exact if and only if for all $A$-modules $N$, the   sequence  $$0\longrightarrow \operatorname{Hom}(M'', N)  \overset{\overline{v}}{\longrightarrow}\operatorname{Hom}(M,  N)\overset{\overline{u}}{\longrightarrow}\operatorname{Hom}(M', N)  \tag{2} $$ is exact.   Here, $\overline{v}$ is the map defined by $\overline{v}(f)=f\circ v$ for every $f\in\operatorname{Hom}(M'', N)$ and $\overline{u}$ is defined likewise.   The proof one of direction, namely $(2)\Rightarrow (1)$ is given in the book, which I am having some trouble understanding. So assuming (2) is exact sequence, the authors remark that "since $\overline{v}$ is injective for all $N$, it follows that $v$ is surjective". Could someone explain why this follows?  Given that $\overline{v}$ is injective, we know that whenever $f(v(x))=g(v(x))$ for all $x\in M$, we have $f=g$. I am not sure how we conclude from this surjectivity of $v$.  Thanks!
Should I really pay off my entire credit card balance each month or should I maintain some balance? How to get text in between two sections? I want to get text in between START &amp; STOP. Remaining text should be deleted, like some_text_1 and some_text_2.  some_text_1 some_text_1 . . . START --- Values that I want to get ... STOP some_text_2 some_text_2   From my findings, sed does similar job, but only oposite, like: sed '/START/,/STOP/d' file which would give me everything else then content in between START &amp; STOP.  Any idea? Recall the following relevant definitions. We say that   $b$ is divisible by $a$ in $R$, or $a\mid b$ in $R$, if $b = r a$ for some $r\in R$. $a$ and $b$ are associates  in $R$ if $a\mid b$ and $b\mid a$, (or, equivalently, if $aR = bR$). $u\in R$ is a unit if it has a multiplicative inverse (a $v\in R$ such that $uv=vu=1$). $a$ and $b$ are unit multiples in $R$  if $a = ub$ for some unit $u\in R$.   Given these definitions, my question is,     If $R$ is a commutative ring with unity and $a,b\in R$ are associates in $R$, are $a$ and $b$ unit multiples in $R$?   I was told that this not always true. But I encountered some difficulties in finding a counterexample. How to solve ~(P → Q) : P & ~Q by natural deduction? I'm trying to solve the following by natural deduction:  ~(P → Q) : P &amp; ~Q  It's a trivial problem if identities are used, as can be seen by the following:   {1}      1.   ~(P → Q)                    Prem. {1}      2.   ~(~P ∨ Q)                   1 Mat. Impl. {1}      3.   ~~P & ~Q                    2 De Morgan's {1}      4.   P & ~Q                      3 DNE   Starting from step 2 of the previous proof, it's also fairly straightforward to reach a conclusion using natural deduction; i.e. starting from ~(~P ∨ Q), which is an identity of ~(P → Q):  ~(~P ∨ Q) :  P &amp; ~Q   {1}      1.   ~(~P ∨ Q)                   Prem. {2}      2.   ~P                          Assum. {2}      3.   ~P ∨ Q                      2 ∨I {1,2}    4.   ~(~P ∨ Q) & (~P ∨ Q)        1,3 &I {1}      5.   P                           2,4 RAA {6}      6.   Q                           Assum. {6}      7.   ~P ∨ Q                      6 ∨I {1,6}    8.   ~(~P ∨ Q) & (~P ∨ Q)        1,7 &I {1}      9.   ~Q                          6,8 RAA {1}      10.  P & ~Q                      5,9 &I   I also found that it's fairly easy to conclude ~Q:   {1}      1.   ~(P → Q)                    Prem. {2}      2.   Q                           Assum. {3}      3.   P                           Assum. {2,3}    4.   P & Q                       2,3 &I {2,3}    5.   Q                           4 &E {2}      6.   P → Q                       3,5 CP {1,2}    7.   ~(P → Q) & (P → Q)          1,6 &I {1}      8.   ~Q                          2,7 RAA ... {1,2}    12.  P & ~Q   In spite of how easy it looks, I can't figure out how to do it. I believe it can be done in 12 steps. Thanks! Show that $S = \sqrt{S^2}$ is a biased estimator of $\sigma$ given a random sample from a normal distribution ... Suppose $Y_1, \ldots, Y_n$ is a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$. Let $S^2$ be the sample variance, which is unbiased for $\sigma^2$.  GOAL: Show that $S = \sqrt{S^2}$ is a biased estimator of $\sigma$.  I realize that $(n-1)S^2/ \sigma ^2$ is chi-square with $(n-1)$ degrees of freedom.  I am given that $$E(S) = E\left[\frac{\sigma}{\sqrt{n-1}} \left[\frac{(n-1)S^2}{\sigma^2}\right]^{1/2}\right] = \frac{\sigma}{\sqrt{n-1}} \int_0^\infty v^{1/2} \frac{1}{\Gamma[(n-1)/2] 2^{(n-1)/2}} v^{(n-1)/2} e^{-v/2} \, dv$$  Can anyone help me understand why they have $v^{1/2}$ and $v^{(n-1)/2}$ (combining them $v^{n/2}$ as I think it should be $v^{1/2}$ and $v^{\frac{(n-1)}{2}-1}$ (combining both terms give $v^{(n-2)/2}$ since the pdf for chi-square distribution is $$ f(v) = \frac{v^{(x/2)-1} e^{-v/2}}{2^{x/2}\Gamma(x/2)},$$ where $x$ is degree of freedom?  Could anyone please tell me which one is the correct answer with explanation?  Thank you I am using pgfplot to draw graphs in latex. Now this is the problem I am facing:  \begin{tikzpicture}[scale=1] \begin{axis}[ybar,grid=major,bar width=2pt,width=8cm, height=8cm,xmin=1,xmax=540,ymin=1,ymax=240] \addplot [draw=black,fill=green]  table [x=size, y=cluster, col sep=comma] {sim-0.4-0.6.csv}; \end{axis} \end{tikzpicture}   It generates the following graph:    Since, I have values mainly between 0-50 on x-axis, I want to zoom there. After that, values comes around 500. S0, there are no graph points between 50 - 450. How should I do that? Vertical line for the the right margin in Notepad++ Number of solutions of $x_1+2x_2+\cdots+kx_k=n$? Suppose $n$ be a given positive integer. Then the Diophantine equation $x=n$ has only $1$ solution. Just by inspection, I found that the Diophantine equation $x+2y=n$ has $\left\lfloor \dfrac{n}{2}+1\right\rfloor$ non-negative solutions for $(x,y).$ Also, according to  the Diophantine equation $x+2y+3z=n$ has $\left\lfloor \dfrac{n^2}{12}+\dfrac{n}{2}+1 \right\rfloor$ non-negative solutions for $(x,y).$  Is there any closed form for the number of non-negative integer solutions for $(x_1,x_2,\cdots,x_k)$ of $$x_1+2x_2+3x_3+\cdots+kx_k=n$$ for a given $k\in\Bbb{N}$? How can I prove these formulas rigorously?  EDIT After a very tedious calculation I found that the equation $w+2x+3y+4z=n$ has  $\left\lfloor \dfrac{n^3}{144}+\dfrac{5n^2}{48}+\dfrac{(15+(-1)^n)n}{32}+1 \right\rfloor$ solutions. This solution completely agree with the approximation given by Rus May. However still I believe that we can do something more than this. Is the uncertainty principle a property of elementary particles or a result of our measurement tools? Resize partition without affecting other OSes in dual-boot I wanted to resize (increase or decrease the partition size) the existing Ubuntu partition without affecting it and the other OS(es) installed in the machine. How can I change the page URL without refreshing the page? Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper     Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$   and find the sum if it exists.   I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you. Could a constant function be considered increasing? Given the definition of an increasing function as a function $f:E\to\mathbb{R}$, where $E$ is nonempty such that $x_1,x_2\in E$ and $x_1&lt;x_2$ imply $f(x_1)\leq f(x_2)$, then couldn't, for example, $f(x)=-1$ considered increasing? Since the inequality is not a strict inequality. Prove that inequality $\sum _{cyc}\frac{a^2+b^2}{a+b}\le \frac{3\left(a^2+b^2+c^2\right)}{a+b+c}$ Let $a&gt;0$,$b&gt;0$ and $c&gt;0$. Prove that:    $$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$     $$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)$$  $\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)$  $\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2$  $\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2$  $\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)$  I can't continue. Help me Mysterious cord running from wall. Safe? I want to be able to ask a question by sending an email to some special email address for Stack Overflow. I have found this convenient in other situations like Google Groups, blogging, and Facebook walls, and it seems that technically it shouldn't be too hard to do.  Does this feature exist already? I'm kind of afraid to ask for features as some of them get downvoted simply since users think they are not neccessary (and not because the post is bad). What is the smallest (installed size) ubuntu based distribution? For ridiculously small hard drives (8GB), what Ubuntu Based (although not necessarily official) distribution offers the smallest installed size?  Let me clarify: I still want a certain amount of personal file space, preferably a few gig. I also don't have a problem with performance, while it is mediocre, a 2.6GHz single-core and 1GB of ram suffice to run ubuntu fine, but i find that it  uses 6-7GB of space. I am aiming for closer to 4GB. On every bootup it's the same:  /dev/sda1: clean, 908443/38690816 files, 44176803/154733312 blocks   Is it some kind of option Ubuntu uses to ensure filesystem consistency or is there something wrong with my HDD? fsck takes up to 30s while booting and so about triples the time needed otherwise.  Full output (partly in German):  Begin: Loading essential drivers ... done. Begin: Running /scripts/init-premount ... done. Begin: Mounting root file system ... Begin: Running /scripts/local-top ... done. Begin: Running /scripts/local-premount ... done. Begin: Running /scripts/local-bottom ... done. done. Begin: Running /scripts/init-bottom ... done. fsck von util-linux 2.20.1 /dev/sda1: sauber, 908443/38690816 Dateien, 44176803/154733312 Blöcke udevd[623]: unknown key 'SYSFS{idVendor}' in /lib/udev/rules.d/45-libticables.rules:6  udevd[623]: invalid rule '/lib/udev/rules.d/45-libticables.rules:6'   * Starting mDNS/DNS-SD daemon                                                 [ OK ]  * Starting Reload cups, upon starting avahi-daemon to make sure remote queues are populated                                                                   [ OK ]  * Starting configure network device security                                  [ OK ]  * Starting bluetooth daemon                                                   [ OK ]  ####* Starting all other stuff Is my US visa still valid after change in name? I'm planning to fly from Ulaanbaatar to Melbourne via Seoul (and Kuala Lumpur).  However, can I actually enter South Korea using this visa waiver? I'm hoping to meet a friend there for a few hours before flying. Also, would traveling via Malaysia affect this policy?  Transit tourists bound for another country Eligible countries  Applicable to all countries (save for Macedonia, Cuba, Syria, Sudan and Iran) that are not granted visa-free entry into Korea.  Applicable to…  Those with a visa (excluding Japanses Group Visa, excluding e-visa) or a re-entry permit for U.S., Japan, Canada, Australia, or New Zealand (5 countries) and traveling to any one of those countries through Korea.  Those traveling from any one of the five countries mentioned above, whose final destination is not Korea.  However, please note that the first bullet points are not applicable to those travelling to/from Guam and Saipan, which are unincorporated territories of the United States.   Requirements  Must have a confirmed onward flight ticket for departure within 30 days after entering Korea, in addition to having no record of criminal offence in the five countries mentioned above.