Does there exists  two real valued functions $f$ and  $g$ on $R$ such that  $f \circ g  = x^{2018}$ and $ g\circ f  = x^{2019}$ ?  My attempt : since  $g \circ f $ is bijective thus  $f$ is one one and $g$ is onto. Now  $f \circ g(-x) = f \circ g(x) $ imply $ g(-x) = g(x) $ (because $ f $ is one one) thus g is even function.  Now i dont know how to proceed from here any hint will be helpfull for me.... (I know no such map exists as answer)
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I have tried a little bit to solve the problem which goes as follows:   My intuition says that there exist no $f:\Bbb R\to\Bbb R$ such that $$f(g(x))=x^{2018}\text{ and }g(f(x))=x^{2019}.$$  Note that $$f(g(f(x)))=f(x)^{2018}\implies f(x^{2019})=f(x)^{2018}$$  Similarly, $$g(x^{2018})=g(x)^{2019}$$ Putting $x=1$ in $f(x^{2019})=f(x)^{2018}$, we get $f(1)=f(1)^{2018}$ and thus $f(1)=0$ or $f(1)=1$.  Similarly, putting $x=1$ in $g(x^{2018})=g(x)^{2019}$ we get $g(1)=g(1)^{2019}$ and thus $g(1)=0,1,-1$.  Now, I can't proceed further. Can anybody solve it? Thanks for assistance in advance.
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How do I show that if there are functions $f,g$ such that$$ f(g(x)+g(y))=bx+cy $$holds for all $b,c\in\mathbb{R}$, then we necessarily have $b=c$? Is this even true? It seems so, but I'm just not sure how to explain it in detail and using theory from functional equations. Any ideas?  Extra: In addition, I'd like to prove that there are no pair of functions $f,g$ such that$$ y^2+z^2=f(x,g(y-x)+g(z-x)). $$ Any ideas on how to prove this? Plus, does anyone know some literature regarding this type of problems?
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Consider $f:A \to B$ and $g: B \to C$. If I know that both $f$ and $g$ are injective, then can I state that $$g  \circ f:A \to C$$ is injective?
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Let $f\colon S^1\to S^1$ be any smooth map. Prove that there exists a smooth map $g\colon\mathbb{R}\to\mathbb{R}$ such that $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$ and satisfying $g(2π)=g(0)+2\pi q$.  The book told me to show that $g(2\pi)=g(0)+2\pi q$, then extend $g(t+2\pi) =g(t)+2\pi q$.  But I don't understand what they mean. How can I get function $g$ from $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$, and what is $g$?
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Here are two sets which I need to define a bijection function in order to show they are equivalent: $$ (A^B)^C \triangleq \{f:C\rightarrow \{g:B\rightarrow A\} \} $$ $$ (A^{B \times C}) \triangleq \{ h:B \times C \rightarrow A\} $$ where $A, B, C$ are sets, and $f, g, h$ represent functions.  I am trying to define a function $F : C \rightarrow A$ , but still struggling with its formal description, and how to use functions like $g$ above.  What is the right way of showing that such bijection $F$ exists?
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Let  $f : \mathbb{R}^2\rightarrow \mathbb{R}$ be a continuous map such that $f(x)=0$ only for finitely many values of $x$. Which of the following is true?   Either $f(x)\leq 0$ for all $x$ or $f(x)\geq 0$ for all $x$. the map $f$ is onto. the map $f$ is one one. None of the above.   What I have done so far is :   I would take polynomial in two variables.. This need not be like $f(x)\geq 0$ for all $x$ or $f(x)\geq 0$ for all $x$.So,first option is eliminated. The map is not one one assuming that $f$ has more than one zero. So, third option is wrong.   I could not think of an example in which it is not onto..  Only examples i am getting in my mind are polynomials and they are onto..  So, I am having trouble with surjectiveness of the function.  Please help me to clear this.  Thank you..
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Suppose $f: A \to B$ and $g: B \to C$ are both surjections.  Since $f$ is surjective, then for every $b \in B$ there exists $a \in A$ such that $f(a)= b$.  Since $g$ is surjective, for every $c \in C$, there's  $b \in B$ such that $g(b) = c$. Then $g(f(a)) = g(b) = c$.   Would this work?  edit; I didn't realize there are similar questions on this site. Don't mind if this post is deleted.
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We have the sets $X,Y$ and $f:X\to Y, \ g: Y\to X $      Prove if $f$ has a right inverse function: $f\circ g=id_Y$ $\iff$ $f$ is onto $Y$ (surjective).   $\Leftarrow \forall y\in Y : \exists x\in X$ then $f(y)=x$ I need to show that $f\circ g(x)=x$ but I don't know how.  I have no clue on to start in the other direction.   BTW, I found below similar question but they're either not quite the same as this one or I didn't understand the answers:
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I've read that function composition "generally does not commute."  Not counting compositions involving the identity function, and compositions of a function and its inverse, are there examples of functions on the reals (for example) $f, g$ where $fg = gf$ outside of these cases?
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How can one show that if $g(f(x))$ is injective and $f$ is surjective then $g$ is injective?  Here is my attempt:  $g(f(x))$ is injective, so  $$g(f(a))=g(f(b)) \iff f(a)=f(b).$$  $f$ is surjective, so $f(x)=y$.  So $y=y'$ such that $f(b)=y'$.  So $g$ is injective.  Is my proof correct?
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There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$, what I tried is if possible then define $g(z)=f(z)-{1\over 2^{1\over z}}$ Then $g({1\over n})=0$ and so $g(z)$ is entire and its $0$ set has limit point in it and so $f(z)={1\over 2^{1\over z}}$ which is not analytic at $0$? Please help!  Edit: OOps! the way I defined $g(z)$ that is not entire! could any one give me hint?
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How would you prove the  bijective? I only know how to prove a bijection by showing (1) If $f(x) = f(y)$, then $x=y$ and (2) There exists an $x$ such that $f(x) = y$  How would you show that for a function like the Cantor pairing function?
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This question is from an exercise in Way of Analysis (section 10.2.4 problem 20).  If $f: \mathbb{R} \rightarrow \mathbb{R}$ is $C^k$ and $f$ is even, then show $F: \mathbb{R}^n \rightarrow \mathbb{R}$ given by $F(x) = f(|x|)$ is also $C^k$.  $C^k$ just means the function has $k$ continuous derivatives, and $|x|$ is referring to the regular Euclidean norm.  I am not sure how to use the fact $f$ is even in the proof and having trouble for a general $k$. Any hints, even on a particular case like $k=2$, are welcome.  My workings:  As a comment suggests $f(x) = F(x)$ for $n = 1$. For $n = 2$, $F(x) = f(\sqrt {x_1^2 + x_2^2}).$ I'm stuck on where to proceed from here. Is there some form of induction here that I am not seeing?
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About a bijection between $\mathbb R$ and $\mathbb R \cup \{\infty\}$  If we'll take a line $f(x)=\begin{cases} ax &amp;, x\in\mathbb R\\ 0 &amp;, x=\infty  \end{cases}$ then we'll have a bijection.   But about this function: $g(x)=\begin{cases} x &amp;, x\in\mathbb R\setminus \{0\}\\ 0 &amp;, x=\infty  \end{cases}$ ? it appears to be a bijection and it's even simpler.
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For nonempty sets $A$ and $B$ and functions $ f: A \to B$ and $g: B \to A$, suppose that $ g \circ f = i_A$. Prove that if $f$ is onto, then $g$ is $1-1$.   Here is what I started, I don't think it's correct.   Suppose $f$ is onto  i.e for each element $b \in B$ there is an element of $a\in A$, $f(a)=b$. (definition of onto).   Let $y,\ z$ be any two elements of $B$.Then $y=f(a)$ and $z=f(b)$ for some $a,b \in A$.   If $g(y)=g(z)$, then $g(f(a)) = g(f(b))$,which implies that $a=b$, so $f(a)=f(b)$, so $y=z$.   Thus, whenever $g$ maps $y$ and $z$ to the same element in $A$, $y=z$, so $g$ is one-to-one.   PS: New to proof writing, any constructive criticism is much appreciated.
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I have two function $f$ and $r$, such that: $y = f(x)$, $r(y) = x$. I read that it means that: $f \circ r = I_{U}$, where $x \in U, y \in U$, where $U$ is a finite set. What does $f \circ r = I_{U}$ mean? What does this circle mean and how could i read all this writing $f \circ r = I_{U}$ ?
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I'm having trouble with this functional equation, to find all functions $f:\mathbb R \to \mathbb R$ for which the following is valid for all $x,y\in \mathbb R$: $f(x) + f(x+f(y)) = y + f(f(x) + f(f(y)))$ I haven't been able to pinpoint any solution, nor to prove surjectivity or injectivity. So 0 progress. Tried with $x=y=0, x=0, y=0, x=-f(y), y=f(x)$, and even with some assumptions on bijectivity, but nothing helped. I don't know the source but seems like a pretty difficult problem, maybe even impossible to solve.