In this case, if the composition is injective then it has to be that the "outer" function is injective, and that if the functions are surjective then their composition is as such.Showing a specific case is a valid method for disproving a claim, as it shows that at a certain time the properties hold but the conclusion is false.Let $A,B,C$ any sets,polo ralph lauren Outlet, and $f,g$ functions as required,alviero martini online. To show that $g\circ f$ is surjective we want to show that every element of $C$ is in the range of $g\circ f$. Assume $c\in C$ then there is some element $b\in B$ such that $g(b)=c$. Since $f$ is surjective we have some $a\in A$ such that $f(a)=b$. It follows that $g\circ f (a) = c$ and therefore the composition of surjective functions is surjective.Now,http://www.itborsealvieromartini.info, consider this proof. There was nothing to limit ourself,Le pliage LongChamp. I did not require anything from $A,B,C$ other than them being sets, and I did not require anything from $f,g$ other than them being functions from/onto the required set in the premise of the claim. Furthermore, I did not check the surjective-ness of the composition by choosing a certain element. I picked and arbitrary element, from a set which was also arbitrary. Using this sort of argument assures us that the property is not dependent on any of the characteristics of the set (for example,Borsa LongChamp, some things are only true on finite,longchamp borse itlay, or infinite, or other certain sets). This allows us to have a very general theorem. Whenever we have three sets,ralph lauren borse, and two functions which are surjective between them,borse alviero martini, then the composition is also surjective!Hold on a second. For $n=3$ we have that $3^2 = 9$, but $2\cdot 3 = 6$, and $6\neq 9$,LongChamp Outlet. So the statement isn't valid after all! Can you see what went wrong,alviero martini outlet?The same thing applies for your statements. Instead of proving them in one particular case, you should proof them for every possible $f$ and $g$ you can think of.[edit]Here's some help with case (b), I suggest you stop reading after every step and try to complete the proof yourself. If it doesn't work,LongChamp Borse, read the next hint.
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