Prove a mapping $C(X,Y)\to \operatorname{Hom}(C(Y), C(X))$ is surjective
Homomorphism between compact Hausdorff spaces.
Cycles final render pitch black? In solid mode with Material or Texture, it is shown fine.   But in the Render mode, everything is pitch black. I have a sun, and the Opacity of the scene is set to 100%.    I am new to using Blender, so I have no idea what is wrong. If you could help, I would appreciate it. If you need more information to help me, I would be glad to give it. I have a chromebook series 5. I want to install Ubuntu onto a USB stick. When I downloaded Ubuntu onto my USB stick, it won't start up.  Do I first have to download Ubuntu onto my chromebook hard drive, open Ubuntu and then make USB stick bootable? And then return back to chrome?  Or is there a way to download directly onto USB and make the stick bootable?  I appreciate any help. How to use stars and bars for solving number of integral solutions to an equality? Prove that $R$ is a commutative ring if $x^3=x$ Let $R$ be a ring satisfying: $\forall x\in R, \; x^3=x$. Prove that $R$ is a commutative ring. Wielding two weapons without "Two-Weapon Fighting" or "Dual Wielder"? We have a Barbarian on our team that very much likes to switch up his weapons; he goes from a Dwarven Hammer, to a pair of Hand-Axes, to occasionally just going bare-fisted. No issues there, he's fairly adept with all 3.  However, I have only just recently picked up on the way that he rolls his attacks for the two Hand-Axes: He rolls the first attack, adds his modifiers etc. Then he rolls the second attack, this time without modifiers, which is all well and good. But I'm not sure if this action uses his bonus action or not, or is considered a separate action (though, being a level 4 Barbarian, he only has one action per turn), or is merely part of the same action.  If we have a look at the Two-Weapon Fighting feature from the Fighter's fighting styles:     Two-Weapon Fighting   When you engage in two-weapon fighting, you can add your ability modifier to the damage of the second attack. (PHB5e p.72)   Similarly, the Dual Wielder feat also uses the Two-Weapon Fighting feat:     Dual Wielder   You master fighting with two weapons, gaining the following benefits:      ...         You can use two-weapon fighting even when the one-handed melee weapons you are wielding aren’t light. (PHB5e p.164, emphasis mine)      So; yes, this gives a bonus to the second weapon attack, but having asked a  to clarify which action this secondary weapon attack belongs to, it was determined that this uses your bonus action;     [...] the second attack requires a bonus action   So after that long winded explanation, here is the question: Without any additional feats, can you wield and use two weapons at once, and if so, how many actions does it require to make a full attack with both weapons in a turn? Another domain name pointing to my server IP? I was checking out my reverse-DNS entry of my domain the other day which is run from a dedicated server.   I noticed that another person/company had their domain pointing to my server? I have no idea who this company is or why they are pointing their domain at my server IP?  How can I go about stopping this or forcing them to change? Show that $\lim_{x \to +\infty}\left(f(x)+f'(x)\right)=0 \Rightarrow \lim_{x \to +\infty} f(x)=0$ How to show that $\lim_{x \to +\infty}(f(x)+f'(x))=0 $ implies $\lim_{x \to +\infty} f(x)=0$? continuous function: preimage of closure of a set is closure of preimage LyX - undefined control sequence but package seems to be added I am using LyX with MikTeX and I am trying to use the following command \oiiint but I am getting the following error:   \oiiint  The control sequence at the end of the top line of your error message was never \def'ed. If you have misspelled it (e.g., `\hobx'), type `I' and the correct spelling (e.g., `I\hbox'). Otherwise just continue, and I'll forget about whatever was undefined.   I have added \usepackage{mathdesign} to the preamble in:  Document -&gt; Settings -&gt; LaTeX Preamble  Why is this not working? How do I beat Uprising on Legendary difficulty? The achievement "Distinguished Service" requires that you beat Uprising on the regular gamemode on Legendary difficulty.  This means we can't pick our usual huge DPS heroes like Bastion and Orisa, and are stuck with Torbjorn, Reinhardt, Mercy, and Tracer.  The phases are as follows:   Hack all 3 terminals Defend the Payload until it's ready Escort the Payload until the gate Destroy all 4 Orisas.   So I ask this after having failed Uprising on Legendary dozens of times.   How do I beat Uprising on all 4 phases?   Specifically asking for the latter 3 phases.    Defending the payload is hard when it gets to the Bastions that drop in because Reinhardt's shield gets melted very quickly and Torbjorn has little to no DPS.  Tracer must be relied on but it seems there is no way to stop the perfect aim of Bastion as strafing around does nothing.  If and when we actually do manage to defend it, we're stuck on escorting, which is even harder thanks to the tank mode Bastions.  And finally, even if we do make it to the final fight, which has only happened once, the fight is so tough that even if we slowly inch ourselves closer to the end we run out of time.  The previous question I asked was different because Uprising (All Heroes) allows you to pick any 4 heroes to form your squad, but the original gamemode will not let you pick. Also, the achievement requires you to play the original mode -- Uprising (All Heroes) on Legendary does not reward players with the achievement. If I type something like this:  {\itshape\LARGE   Dunt venim dolorerosto do odit   ametum vulla conum dolore   conulput in vero od el   Dunt venim dolorerosto do odit   ametum vulla conum dolore   conulput in vero od el }   it results in a block of 4 lines with narrow spacing between the 1st and 2nd line and between the 3rd and 4th line. Line spacing between the 2nd and 3rd line is normal.  I can't fully explaing it to myself so I am looking for a short description for this behaviour. I know for example that with a \par at the end of the text (inside the group), the calculation would correct. But I have no idea why... Numbered list in rows in tables I want to have numbered listing in each row, I've tried with \rownumber, but it doesn't work. I wrote a number at the beginning, but Could it be automatically?  Can you help me? This is my code:  \documentclass[10pt, a4paper]{article}  \usepackage{fontspec} \setmainfont{Arial}  \usepackage{color, colortbl} \definecolor{Gray}{gray}{0.9}  \begin{document}  \begin{table}[h!] \tiny \begin{tabular}{|&gt;{\raggedright\arraybackslash}m{30mm}|&gt; {\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|&gt;{\raggedright\arraybackslash}m{5mm}|} \hline \multicolumn{1}{|&gt;{\centering\arraybackslash}m{30mm}|}{\rowcolor{Gray}\textbf{Text}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 1}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 2}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 3}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 4}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 5}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 6}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 7}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 8}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 9}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 10}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 11}} &amp; \multicolumn{1}{|&gt;{\centering\arraybackslash}m{5mm}|}{\textbf{Sem 12}}\\ \hline 1. Text text text text text text text text text text text text text &amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;\\\hline 2. Text text text text text text text text text text text text text &amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;\\\hline 3. Text text text text text text text text text text text text text  &amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;\\\hline 4. Text text text text text text text text text text text text text &amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;\\\hline  \hline \end{tabular} \end{table}  \end{document} Windows 10 - Users already logged in at boot probability that there is at least one defective part Multiple file upload in php Why doesn't the "zig-zag" comb deformation retract onto a point, even though it's contractible? I am starting to read Hatcher's book on Algebraic Topology, and I am a little stuck with exercise 6(c) in Chapter $0$. Unfortunately a picture is involved so it doesn't quite make sense for me to repeat it here - but I'll do it anyways (maybe it helps)     Let $Z$ be the zigzag subspace of $Y$ homeomorphic to $\mathbb{R}$ indicated by the heavier line in the picture:       Show there is a deformation retraction in the weak sense of $Y$ onto $Z$, but no true deformation retraction."   Now the definition for deformation retraction and weak deformation retraction as as follows:  We say that $f: X \to X$ is a deformation retraction of a space $X$ onto a subspace $A \subset X$ if there exists a family of maps $f_t : X \to X$ with $t \in [0,1]$ such that $f_0 = \mathbb{1}$ (the identity) and $f_1 (X) = A$ and also $f_t$ restricts to the identity on $A$ for each $t$.   A weak deformation retraction is almost the same, only that we now relax the conditions $f_1(X) = A$ to $f_1 \subset A$ and, for each $t \in [0,1]$ we require that $f_t(A) \subset A$.  Hence, as far as I understand the question and the concepts involved, I need to show that basically no map can be constructed such that the bold zigzag - line stays put while the thin lines retract to the bold line over a finite time interval $0 \leq t \leq 1$. However, I should be able to show that I can pull the thin lines continously onto the thick line, provided I am allowed to move points on the zigzag line .. is that correct ?  Now, the problem is I can't see why I shouldn't be able to do the former, given that I could find a weak deformation retract. Any help would be great! The locker problem - why squares? There are 1000 lockers in a high school with 1000 students.  The    problem begins with the first student opening all 1000 lockers; next    the second student closes lockers 2,4,6,8,10 and so on to locker 1000;   the third student changes the state (opens lockers closed, closes    lockers open) on lockers 3,6,9,12,15 and so on; the fourth student    changes the state of lockers 4,8,12,16 and so on. This goes on until    every student has had a turn.      How many lockers will be open at the end?   I worked out this problem by doing it for 10 lockers. The pattern was obvious (1,4,9 remained open).It's easy to then say that 31 lockers remain open at the end. I don't quite understand why only square numbers remain open at the end. Is there some deeper reason? Difference between == and === in JavaScript What is the difference between == and === in JavaScript?  I have also seen != and !== operators.  Are there more such operators? error with inner class object fontenc vs inputenc