I  don't know what is the example of cochain $ f$ from $A^*$ to $B^*$ such as the map $f^i$  from $A^i$ to$ B^i $ is bijective.
I am looking for a injective cochain map $\psi: C^*\rightarrow D^*$ such that the map $\Psi^i$ from $C^i$ to $D^i$ is injective, but the map $\psi^*$ from  $H^i(C^*)$ to $H^i(D^*)$ is not injective for any $i\geq0$.
When an air-filled balloon is released without its opening tied up, it moves in a circular path rather than a straight line. Why is that? Equality in an integral estimate Let $u,v\in W^{1,p}(\Omega)$ with $1&lt;p&lt;\infty$ and $\Omega$ be a bounded smooth domain in $\mathbb{R}^N$. Suppose, we have $$\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla v\,dx=\left(\int_{\Omega}|\nabla u|^p\,dx\right)^\frac{p-1}{p}\left(\int_{\Omega}|\nabla v|^p\,dx\right)^\frac{1}{p} $$ Suppose the equality $$ \int_{\Omega}|\nabla u|^p\,dx=\int_{\Omega}|\nabla v|^p\,dx $$ holds. Then can we say that $u=v$ in $\Omega$? I tried in the following way: Since the first given equality, which is the equality in Holder's inequality holds, and by the second given equality holds, we have $|\nabla u|=|\nabla v|$ in $\Omega$. Then I am unable to proceed. Can someone please help to give an argument? How can I edit this trigger to only fire once? The trigger creates a related record when a checkbox is updated. The problem is it is creating a new related record everytime the record is edited. I want to edit the trigger to only create the related record once. (When the criteria is first met). The code is below  Trigger copyEquipmentonsite on Unit_Placements_Sales__c(after insert, after update) {      List&lt;Services__c&gt; sub=new List&lt;Services__c&gt;();      for(Unit_Placements_Sales__c u : Trigger.new) {           if(u.Service_Month_is_Next_Month__c == TRUE) {                 Services__c s = new Services__c();                 s.Equipment_Onsite__c=u.ID;                 s.Equipment_Type__c=u.Equipment__c;                 s.Machine_Specific_Notes__c=u.Machine_Specific_Notes__c;                    s.Product__c=u.The_Product__c;                   s.Dell_Tag_Number__c=u.Dell_Tag_Number__c;                 s.Serial_Number__c=u.Serial_Number__c;                  s.XD_Number__c=u.XD_Number__c;                 sub.add(s);           }           if(sub.size()&gt;0)           insert sub;      } } Broken screen while debug mode was disabled. How can I re-enable adb? The screen of my phone is physically broken, so it doesn't display anything, but the touch aspect does seem to be working, as I can unlock my phone, and get haptic feedback.  Before I go to get it replaced, I'd like to back up some personal data from the built-in memory to the SD card. I'm fairly certain that if I can get adb to recognize the device, then I can use  to see where I am touching the screen and do what I need to do. Problem is, when I plug it in, it's not recognized, while my emulator is.  I suspect that I might not have USB debugging enabled, which I think is necessary for adb to work. If this is the case, can someone give me screenshots of their Droid X (running 2.3.X) going from the home screen, through all the menus, to activating debug mode? Hopefully I can do it by dead reckoning and adb from there.  Edit: I haven't changed anything about the ROM or launcher; it's all stock. How can I change the umask for one command only? If you have such a good way to do formatting in questions/answers why isn't it used for the "about me" section on the profile page?  It looks easy to make and adds just a little bit more consistency. How do I make a Villager Trade a Spawn Egg I am trying to make a Custom Trade on a villager. I need to make him buy 8 gold nuggets and sell a Cow Spawn Egg.  basically you pay 8 nuggets for a Cow Spawner.  The ID of the Cow Spawn Egg is 92 but I don't know where to put it in the command /summon Villager ~ ~2 ~ {Invulnerable:1,CustomName:Spawner_Vendor,Profession:2,Career:1,Offers:{Recipes:[{buy:{id:gold_nugget,Count:8},sell:{id:spawn_egg,Count:1}}]}} Isomorphism $f$ preserves the order of an element? Let's say $f$ is an isomorphism $f:G  \rightarrow G'$, where $G$ and $G'$ are multiplicative groups.  Then for $x\in G$, if $f(x) = x' \in G'$. Do we have always have $\text{ord}(x) = \text{ord}(x')$? Why? How do I view my whole character? I have played Oblivion and in Oblivion you can see your person on the menu screen. When I go into 3rd person all I see is their back. Is there any way I can see my whole character? Let $\alpha(t)$ be a regular curve. Suppose there is a point $a$ in $R^3$ such that $\alpha(t)-a$ is orthogonal to $T(t)$ for all t. Prove that $\alpha(t)$ lies on a sphere.  So, I let $\alpha(t)=(x(t), y(t),z(t))$ and $a=(a,b,c)$.  Since, $\alpha(t)-a$ is orthogonal to $T(t)$,  $&lt;\alpha(t)-a,T(t)&gt;=0$.  But I don't know how to solve this.. Cyclic vector implies commuting linear operator is a polynomial Let $T$ be a linear operator on the finite dimensional vector space $V$. Suppose $T$ has a cyclic vector. Prove that if $U$ is any linear operator which commutes with $T$, then $U$ is a polynomial in $T$. Boolean and Texts I have used Hexagon for years but find that it sometimes crashes when trying to do Boolean operations with texts.  I have been trying to make the two tables of the Ten Commandments and have been trying to use Blender.  I did it about two years ago in Blender and lost everything on my computer so I am trying again with the latest version of Blender and am having difficulty.  I start with a cube and then get it close to the dimensions I would think the Ten Commandments might look like. I then bevel the top using the bevel modifier. But in doing it this way I don't have any cross sub divisions on the face of the object. I have tried sub dividing first but then I find that I can't bevel the top arches as far as I would like. I still don't know how to add divisions after I bevel the top. Yes I am still a novice at this program.  In any case I did figure out how to make text and turn them into an object. When I make the texts I do all four of the commandments for one side and then the other six for the other side. So this makes allot of texts to cut out.  When doing the Boolean operation in Difference sometimes nothing happens. I am thinking I need to have some sub - divisions on the face of the sized cube after it is beveled but am not really sure if this is the problem.  This whole thing about normal's is still Greek to me. I have watched some videos on how to cut letters into a cube but for some reason it does not work this easy after I bevel the top of the cube. Or it could be that Blender does not like a bunch of texts all at once.  Could someone show me a video or give me some simple steps to make my project work? If a two vectors $\bf u$ and $\bf v$ in linearly independent, can their images be linearly dependent given they were transformed by a linear transformation $T$? I think not. My reasoning If the images are linearly dependent, then $c_1T({\bf u})+c_2T({\bf v})={\bf 0}$ for scalars $c_1$ and $c_2$, not both $0$. Then $c_1T({\bf u})+c_2T({\bf v})={\bf 0}$ $T(c_1{\bf u}+c_2{\bf v})=T({\bf 0})$ $c_1{\bf u}+c_2{\bf v}={\bf 0}$ Therefore, ${\bf u}$ and $\bf v$ are linearly dependent. My reservation A homework problem seems to contradict the above. Here's the problem: Let $\mathbb R^n\rightarrow\mathbb R^m$ be a linear transformation. Suppose $\{{\bf u},{\bf v}\}$ is a linearly independent set, but $\{T({\bf u}),T({\bf v}\})$ is a linearly dependent set. Show that $T({\bf x})={\bf 0}$ has a nontrivial solution. static variables in an inlined function I have a function that is declared and defined in a header file. This is a problem all by itself. When that function is not inlined, every translation unit that uses that header gets a copy of the function, and when they are linked together there are duplicated. I "fixed" that by making the function inline, but I'm afraid that this is a fragile solution because as far as I know, the compiler doesn't guarantee inlining, even when you specify the "inline" keyword. If this is not true, please correct me.  Anyways, the real question is, what happens to static variables inside this function? How many copies do I end up with? In Walter Rudin's Principles of mathematical analysis Exercise 5.21, it is proved that for any closed subset $E\subseteq \mathbb{R}$, there exists a smooth function $f$ on $\mathbb{R}$ such that $E=\{x\in \mathbb{R}\mid f(x)=0\}$.   Since $E^c$ is open in $\mathbb{R}$, $E^c$ is a countable disjoint union of open intervals $(a_i,b_i)$. On an interval $(a,b)$, if we let $f(x)=\exp(\frac{1}{(x-a)(x-b)})$, $x\in(a,b)$; $f(x)=0$, $x\notin(a,b)$, ($f(x)=\exp(\frac{1}{(x-a)})$ or $f(x)=\exp(\frac{1}{(x-b)})$ on $(a,\infty)$ and $(-\infty,b)$ resp.), then $f$ is smooth. Hence we can get a smooth function on $\mathbb{R}$ such that $E$ is the zero point sets of $f$.  (1). How about a closed subset $E$ of $\mathbb{R}^n$? is it true that for any closed subset $E\subseteq \mathbb{R}^n$, there exists a smooth function $f$ on $\mathbb{R}^n$ such that $E=\{x\in \mathbb{R}^n\mid f(x)=0\}$?  An open set in $\mathbb{R}^n$ cannot be written as a disjoint union of countable open balls hence the proof above is not valid. The partition of unity only claims that for an  open set $U$, there exists open $V$ in $U$ such that $\bar V\subseteq U$ and a smooth function $f$ on $\mathbb{R}^n$ such that $supp f\subseteq U$, $f|_V=1$.   (2). The taylor series of $f(x)=\exp(\frac{1}{(x-a)(x-b)})$ does not converge around points $a,b$ hence $f$ is not analytic at $a,b$. Is it true that for any closed subset $E\subseteq \mathbb{R}$, there exists an analytic function $f$ on $\mathbb{R}$ such that $E=\{x\in \mathbb{R}\mid f(x)=0\}$? Best way to put commas into large numbers I've started a new job and it involves looking at a bunch of big numbers. Is there an easy way to add commas to an int or decimal field to make it readable?  For example, SQL Server outputs the column on the left, but for my own sanity, I need it to look like the one on the right:  2036150 -&gt; 2,036,150     ...or would I have to write some heinous   left(right(vandalized_data),6),3) + ',' + right(left(vandalized_data),6),3)   function?  The perfect thing would be commas in the display grid, then plain integers in the output. Why is diamond transparent while graphite is not? Diamond and graphite are both made of the same atom, carbon. Diamond has a tetrahedron structure while graphite has a flat hexagonal structure. Why is diamond transparent while graphite is not (at least not with more than a couple of layers)? I have heard of some methods, but none of them have stuck. Personally I try to avoid complex types in C and try to break them into component typedef.  I'm now faced with maintaining some legacy code from a so called 'three star programmer', and I'm having a hard time reading some of the ***code[][].  How do you read complex C declarations? I have been thinking about the following formulation of the Axiom of Choice.:     For any relation $R$, there is a function $H$, which is a subset of $R$, and such that $\operatorname{dmn}H=\operatorname{dmn}R$.   At first, it seemed surprising to me that this did not follow from Comprehension (i.e., from the schema  $\forall x_1, x_2, ..., x_n \forall C\exists B \forall x ( x \in B \Leftrightarrow x \in C \land \phi )$, where $B$ does not occur free in $\phi$).    I thought that for any relation, it seems that we should be able to define a subset of it with the same domain meeting a "functional condition".  But when I tried to do this, I couldn't.  However, I felt like I was getting close when I let $B$ be free in $\phi$, i.e., when I let myself say that $w$ is in $R$ and also anything in $H$ meets such and such conditions, where $H$ is the set (function) whose existence I'm trying to prove.    But my understanding is that Choice does not even follow from Naive Comprehension.  So, I wondered if someone could explain the following precisely.   Why does the above formulation of Choice does not follow from regular Comprehension? Why does it not even follow from Naive Comprehension (assuming that it does not)? Do inequalities hold under differentiation? If I have the inequality $f \leq g$, does this imply $f'\leq g'$?  The reason I am asking is because I am using this fact to prove a question with Taylor theorem.