Div won't center horizontally
How to horizontally center an element
Altium: How can I edit a component from a library Prove homotopic attaching maps give homotopy equivalent spaces by attaching a cell Prove: If $f,g:S^{n-1} \to X$ are homotopic maps, then $X\sqcup_fD^n$ and $X\sqcup_gD^n$ are homotopy equivalent.   I think it can be proved by showing they are both deformation retracts of $X\sqcup_H(D^n\times I)$ where $H$ is the homotopy between $f$ and $g$.  However, I have difficult in proving that the deformation retracts are continuous map. In fact, I have difficulty in representing a map in quotient spaces like $X\sqcup_fD^n$. I think a map from $X\sqcup_fD^n$ to $W$ can be represented by two maps: $m_1: X\to W$, $m_2: D^n\to W$, where for $x\in S^{n-1}$, $m_1\circ f(x)=m_2\circ i(x)$.  Then I construct the deformation retract this way: $m_1: X\to X$. For $x\in H(S^{n-1},t)$, $m_1(x)=H(S^{n-1},0)$, otherwise $m_1(x)=x$.  $m_2: D^n\times I\to D^n\times {0}$: $m_2((D^n,t))=(D^n,0)$.  It is easy to verify that $m_1$ and $m_2$ define a map from $X\sqcup_H(D^n\times I)$ to $X\sqcup_fD^n$. As long as this is a continous map, obvioulsy then we find a deformation retract. But it seems such a map is not continous? $D_8$ as a derived subgroup Every undergraduate student knows that there are (exactly) two non-abelian groups of order 8: Dihedral ($D_8$) and Quaternion ($Q_8$). The group $Q_8$ has many interesting properties; simple of them are: it has unique element of order $2$, all subgroups are normal, etc. A property, which is strange and not obvious, the group $Q_8$ has, is that there is no group $G$ such that $G/Z(G)$ is isomorphic to $Q_8$. On the other hand, $D_8$ has no such property; there are (infinitely many) groups $G$ such that $G/Z(G)$ is isomorphic to $D_8$; example - $D_{16}$.   The question, I would like to ask, is related to a property of $D_8$, which is not shared by $Q_8$. There are (infinitely many) groups whose derived subgroup is $Q_8$; example- $SL_2(3)$. For groups $G$ of order &lt;100, one can check on GAP that $D_8$ is not a derived subgroup of any group. The natural question is then:  Question 1: Does there exist a finite/infinite group whose derived subgroup is equal to $D_8$?  Question 2: Does there exist a finite/infinite group whose derived subgroup is equal to (general) dihedral group $D_{2n}$ of order $2n$ ($n&gt;2$)? What is the best way to convert seconds into (Hour:Minutes:Seconds:Milliseconds) time? How can I set the box plot (i.e. axis) ratio of a usual 2D plot to a given value, say 4:3.  I don't want to use unit vector ratio since the quantities of the independent and dependent variables are of completely different orders of magnitude. Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional? It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this? Question about Casella and Berger's proof of MLE invariance In Casella and Berger, p. 320, they have a proof of the invariance of the MLE. Let $g: \theta \mapsto \eta$ be a function. They define the induced likelihood as  $$ L^*(\eta \mid X) = \sup_{\{\theta: g(\theta) = \eta\}} L(\theta \mid X). $$  This ensures that $g$ is one-to-one with respect to the likelihood functions because if $g$ is not one-to-one, there may be multiple values of $\theta$ that map to a given $g(\theta) = \eta$.  Then they note  $$ \sup_{\eta} L^*(\eta \mid X) = \sup_{\eta} \sup_{\{\theta: g(\theta) = \eta\}} L(\theta \mid X) = \sup_{\theta} L(\theta \mid X). $$  The first equality just applies the definition of the MLE of $\eta$. However, the next inequality confuses me. They write,     The second inequality follows because the iterated maximization is equal to the unconditional maximization over $\theta$...   Can someone justify this claim or provide some intuition if the claim is supposed to obviously follow from definitions? Is it ok to answer based someone's older answer on SE? Possible Duplicate:             This question will probably be closed in a few minutes. It is a duplicate but I couldn't find the question using with meta search.  Anyway, , . In this case, is vote to close the new question as a possible dublicate not the best choice? Looks to me a little bit not ethical. I'm using an API that requires me to pass a function pointer as a callback. I'm trying to use this API from my class but I'm getting compilation errors.   Here is what I did from my constructor:  m_cRedundencyManager-&gt;Init(this-&gt;RedundencyManagerCallBack);   This doesn't compile - I get the following error:      Error  8 error C3867: 'CLoggersInfra::RedundencyManagerCallBack': function call missing argument list; use '&amp;CLoggersInfra::RedundencyManagerCallBack' to create a pointer to member   I tried the suggestion to use &amp;CLoggersInfra::RedundencyManagerCallBack - didn't work for me.   Any suggestions/explanation for this??  I'm using VS2008.  Thanks!! Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. Prove that $G $ is an abelian group. Find all functions $f(x)$ such that $f\left(x^2+f(y)\right)$=$y+(f(x))^2$ I have a JavaScript object like the following:  var p = {     "p1": "value1",     "p2": "value2",     "p3": "value3" };   Now I want to loop through all p elements (p1, p2, p3...) And get their keys and values. How can I do that?  I can modify the JavaScript object if necessary. My ultimate goal is to loop through some key value pairs and if possible I want to avoid using eval. How to turn hundreds of text URLs in Excel into clickable hyperlinks? Showing that a group of order $pq$ is cyclic if it has normal subgroups of order $p$ and $q$ This is question 2.9 #9 from Topics in Algebra by Herstein:     If $o(G)$ is $pq$ where $p$ and $q$ are distinct prime numbers and if $G$ has a normal subgroup of order $p$ and a normal subgroup of order $q$, prove that $G$ is cyclic.   Let $N$ denote the normal subgroup of order $p$ and $M$ denote the normal subgroup of order $q$. Here are a couple things I noted while exploring the problem:   $N$ and $M$ are cyclic, since they have prime order. $G = NM$ [1].   Any hints where to look next?    [1] Since $N,M$ are normal, $NM \le G$. Then $o(NM) \mid o(G)$, so $o(NM)$ is either $1$, $p$, $q$, or $pq$. The first three don't work because they force one of $N$ or $M$ to be $(e)$, so we have $o(NM) = pq$. Therefore, $NM = G$. How do we know that certain quantum effects are random? Banner ads should not be shown on the right-hand ad slot still frame camera solve I've found a number of how-tos for solving video but I'm just looking to put an element in a photo and can't seem to find a how-to for that.  Is it possible in blender to do a camera solve with a still image?  Thanks. I'm trying to understand this proof and I can follow it up the the second last paragraph where it states what happens if n is odd/even. I don't understand why there is just one conjugacy class when n is odd and two when n is even.    Help much appreciated, thanks! There's a way to safely try KDE on top of an Ubuntu installation, keeping the possibility to revert the changes in case I dislike how it runs? Why is the mean function in Gaussian Process uninteresting?