Proving a function is surjective given the composition is surjective Here is the question $f:X \rightarrow Y$ and $g: Y \rightarrow Z$ are functions and $g \circ f$ is surjective, is $g$ surjective?  My proof: If $g \circ f$ is surjective than $\forall z \in Z \; \exists x \in X \; \mid (g \circ f)(x)=z  $ Suppose $f$ is surjective, than $\forall y \in Y \; \exists x \in X\; \mid f(x)=y$. By def. of $(g\circ f)(x)=z$ we have $g(f(x))=z$ and since $f(x)=y$ than we have $g(y)=z$ which implies surjectivity therefore $g$ is surjecive.  Is this the correct way to prove that $g$ is surjective? Or can I not assume surjectivity on $f$
Show that if $g \circ f$ is injective, then so is $f$. The Problem:  Let $X, Y, Z$ be sets and $f: X \to Y, g:Y \to Z$ be functions.  (a) Show that if $g \circ f$ is injective, then so is $f$.  (b) If $g \circ f$ is surjective, must $g$ be surjective?  Where I Am:  So, I really have trouble with these, for some reason. I can draw pictures and make sense of the problems, but writing down proofs is very difficult for me.   Basically, for (a), I ended up with some complicated statement involving an implication implying another implication and then tried to derive a contradiction. It just got so convoluted that I couldn't make sense of it anymore, and I know there's a quick, elegant way to show it.  For (b), I know that $g$ need not be surjective. Once again, though, proving it directly from definitions has given me a bit of a headache.  Any help here would be appreciated. Thanks in advance.  The Proofs!  Ok, I did it. Thanks for the help, everyone! Let me know if there's anything wrong with these proofs, or if they could bet any better.  (a) Suppose $f$ is not injective. Then $$ f(x_1)=f(x_2) \implies x_1 \ne x_2 \text{        }(*).$$ Let $f(x_1)=y_0=f(x_2)$ and let $g(y_0)=z_0$. Then $$ (g \circ f)(x_1) = (g \circ f)(x_2) = g(y_0) = z_0. $$ Since $g \circ f$ is injective, $$ (g \circ f)(x_1) = (g \circ f_2) \implies x_1 = x_2. $$ However, this contradicts $(*)$. Therefore, $f$ must be injective.  (b) Suppose $g$ is not surjective. Then  $$ \forall y \in Y, \exists z \in Z \text{ such that } g(y) \ne z \text{         }(**).$$  Since, $g \circ f$ is surjective,  $$ \forall z \in Z, \exists x \in X \text{ such that } g(f(x)) = z \text{       } (***). $$  Let $f(x) = y$. Then,  $$ g(f(x)) = g(y) = z. $$  Because of $(***)$, this is true for all $z \in Z$, which contradicts $(**)$. Therefore, $g$ must be surjective.
Calculating optimal number of bins in a histogram I'm interested in finding as optimal of a method as I can for determining how many bins I should use in a histogram. My data should range from 30 to 350 objects at most, and in particular I'm trying to apply thresholding (like Otsu's method) where "good" objects, which I should have fewer of and should be more spread out, are separated from "bad" objects, which should be more dense in value. A concrete value would have a score of 1-10 for each object. I'd had 5-10 objects with scores 6-10, and 20-25 objects with scores 1-4. I'd like to find a histogram binning pattern that generally allows something like Otsu's method to threshold off the low scoring objects. However, in the implementation of Otsu's I've seen, the bin size was 256, and often I have many fewer data points that 256, which to me suggests that 256 is not a good bin number. With so few data, what approaches should I take to calculating the number of bins to use? My wife has a Nikon D3100 that I've been using to film, but the battery typically won't last through long shoots.  I've gathered from searching around that I can't charge the camera with USB, but I wondered if there's ANY way to do that. It seems at least possible that an adapter could exist that would fill in for the battery and plug into the wall, but I haven't been able to find such a thing.  Is there such a product or technique? It would be nice to not buy a second battery, I'm really hoping for a "set it and forget it" solution. Following an online tutorial that uses an object named "icosahedron" from Add > Mesh > Regular Solids > Archimedean > Icosahedron. This object is not available in my version of Blender (2.74). Presenter asks me to include an add-on named "Add Mesh: Regular Solids", which again is not listed in my User Preferences dialog. What can I do about it? How confident is my model? Is there a way to know? Why does GPS depend on relativity? Lewis Carroll's Pillow Problem via Bayes' Theorem In Pillow Problems and a Tangled Tale Lewis Carroll presents the following problem:     A bag contains a counter, known to be either white or black. A white counter is put in, the bag is shaken, and a counter is drawn out, which proves to be white. What is now the chance of drawing a white counter?   One can easily reason that in the end, the bag must be in one of three states:   Black remains, new white taken New white remains, original white taken Original white remains, new white taken   In two of those states, a white counter remains to be drawn. Therefore, the chance is 2/3.  How can one arrive to this answer using Bayes' theorem? How do electric currents make light? In an incandescent light bulb, electrons flow through wires to the bulb, where is passes through a filament, and the filament lights up.  Does this work because of several high energy electrons losing some of their energy to disturb the electromagnetic field, creating photons? The electrons just leave the filament afterward and are replaced by more high-energy electrons from the negative terminal. Or does it work differently?  As a secondary question, is there a type of light bulb that directly converts electron flow to photons? Why isn’t singular ‘they’ used with 3Sg verb forms? There are many (duplicate) questions about the acceptance, popularity and history of singular they (and their, them and themself) around here, it even got a tag of its own. If I didn’t miss anything, however, the proper verb form hasn’t been questioned yet.  As we all know, English third person singular pronouns (it, she, he and one, +body), names (Alice, Bob, …) and nouns (student, teacher, …) demand the +(e)s suffix be added to the finite verb form in simple present, where some auxiliary verbs have a “special” form (is &lt; *bes / *ares and has &lt; *haves). All other subjects don’t, including plural third person pronoun they.  When the plural you replaced thou (with thee, thy / thine), the other marked verb form that had remained in English – i.e. suffix +(e)st or +t – vanished, too. The first and second persons only ever appear as pronouns (I, we; you), not names or nouns, so there was no strong inclination to keep the verbal inflection.  The second person precedent would suggest that singular they be used with uninflected verb forms which is how it’s usually encountered in the wild. Assuming that they someday replaces he and she (and maybe it) it would lead to disagreement with the words the pronoun stands in for:        Alice         goes to her   place  by herself.   Bob           goes to his   place  by himself.   Alice and Bob go   to their place  by themselves. – (common)   Alice and Bob go   to their places by themselves. – (separate)                  She  goes to her   place  by herself.   He   goes to his   place  by himself.   They go   to their place  by themselves.   They go   to their places by themselves.                  They ?goes to their place  by themself.   ditto   They  go   to their place  by themselves.   They  go   to their places by themselves.                  They ?go to their place  by ?themselves.   ditto   They  go to their place  by  themselves.   They  go to their places by  themselves.      So why doesn’t singular they afford s on present-tense verbs like all other third person singular subjects do?   I learned about singular they only long after I had been taught English as a second language in school. That’s why it’s still a conscious decision to use it and hence I could easily adapt to use s forms with it, but would that sound and look funny / strange / wrong to native speakers? In how many ways can a number be expressed as a sum of consecutive numbers? All the positive numbers can be expressed as a sum of one, two or more consecutive positive integers. For example $9$ can be expressed in three such ways, $2+3+4$, $4+5$ or simply  $9$. In how many ways can a number be expressed as a sum of consecutive numbers?  In how many ways can this work for $65$?  Here, for $9$ answer is $3$, for $10$ answer is $3$, for $11$ answer is $2$. Prove $\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2$ For positive integers Numbers  $a, b, c $ prove that  $$\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq  2$$ This inequality above take long time to prove it and still couldn't complete it.  How I can prove this inequality?  Any hint will help. Thanks Are there surface-mount pin header footprints for KiCAD? Is every Lie subgroup of $SU(2)$ closed? Do I need a visa for Georgia? I am a Nigerian, holding a valid schengen visa issued by the French embassy, and a multiple entry visa for the United States. Will I be allowed to enter Tbilisi in Georgia? Linear independence of $\sin(x)$ and $\cos(x)$ In the vector space of $f:\mathbb R \to \mathbb R$, how do I prove that functions $\sin(x)$ and $\cos(x)$ are linearly independent. By def., two elements of a vector space are linearly independent if $0 = a\cos(x) + b\sin(x)$ implies that $a=b=0$, but how can I formalize that? Giving $x$ different values? Thanks in advance. Tags are not showing up on the right side of the screen This is on Stack Overflow. On Meta I'm able to see what the tags are on the right side of the screen. In fact, I think yesterday or some point earlier today I saw tags on the right side of the screen (I can't screenshot or remember correctly).  It's right above the asked and active heading.  I know this is minor but what happened to it? It might be useful for browsing instead of having to look at the bottom of the question post.    Found an image on Google and it shows the tags on the right side of the screen: In a question related to introductory Galois theory, I was asked to given an example of a tower of fields $F \subset K \subset E$ such that $E$ is a splitting field for some polynomial $f(x) \in F[x]$, but that $K$ need not be a splitting field. The example that comes to mind for me is $\mathbb Q \subset \mathbb Q(\sqrt[3]{2}) \subset \mathbb Q(\omega_3, \sqrt[3]{2})$.  So I know that the minimal polynomial for $\mathbb Q(\omega_3, \sqrt[3]{2})$ over $\mathbb Q$ is $x^3-2$. This polynomial does not split in the intermediate field $\mathbb Q(\sqrt[3]{2})$.  I wanted to make sure that I'm justifying that this intermediate field is not a splitting field correctly. Now I may be confused about the books terminology. When we say something is not a splitting field, we always have to refer to a specific polynomial correct? I've read a term called a normal extension which I think refers to one in which every polynomial splits, but my book doesn't mention these.   If splitting field means with respect to the polynomial $x^3-2$, then I've already demonstrated that $\mathbb Q(\sqrt[3]{2})$ is not a splitting field. Also, it couldn't be a normal extension either because that polynomial doesn't split.  Does it sound like I'm understanding this correctly or are there some subtleties I might be missing? A new Victorian-set British novel begins sentences containing h-dropping with lower-case e's, as in " 'e took my money, 'e did." This seems incorrect but I'm at a loss to find the rule. A particle experiences an acceleration described by  $$ a=kx^{-2} $$ where x is the displacement from the origin and k is an arbitrary constant.  To what value does the velocity v of the particle converge to as x approaches infinity if the particle starts at some point x0?  If I approach this problem with energy, then $$ W = \int F \mathrm{d}x $$ $$  = \int_{x_0}^\infty mkx^{-2} \mathrm{d}x $$ $$  = mk(-\infty^{-1}+{x_0}^{-1}) $$ $$ W = K = mk{x_0}^{-1} $$ $$ \frac{1}{2}mv^2 = mk{x_0}^{-1} $$ $$ v = \sqrt{2k{x_0}^{-1}} $$  How would I solve this problem with pure kinematics? (there appears to be some sort of cyclical dependency where acceleration affects velocity, velocity affects displacement, and displacement affects acceleration)  Likewise, two particles experience accelerations described by $$ a_1=k_1x^{-2} $$ $$and $$ $$ a_2=-k_2x^{-2} $$ where x is the distance between the two particles  What two velocities do the particles reach as x approaches infinity if the two particles are initially separated by some x0? For $f$ an analytic function, what is How do I set OS X to boot by default? I set Windows 7 to my Boot Camp. Windows 7 starts now by default in the normal boot. However, I want that OSX boots by default and Windows 7 only as I press ALT and then select it.  How can you set OSX to boot by default?