Let $X = \mathbb{R}$ and $Y = \left \{ x \in \mathbb{R}\mid x ≥ 1 \right \}$. Define $G : X → Y$ by $G(x) = e^{x^2}$. Prove that $G$ is onto.
122906
Let $X = \mathbb{R}$ and $Y = \{x \in \mathbb{R} :x ≥ 1\}$, and define $G : X → Y$ by $$G(x) = e^{x^2}.$$ Prove that $G$ is onto.  Is this going along the right path and if so how do get the function to equal $y$?  $G: \mathbb{R} \to\mathbb{N}_1$. Let $y$$\in $$\mathbb{N_1}$. claim: $\sqrt{\ln y}$ maps to $y$. Does $\sqrt{\ln y}$ belong to $\mathbb{N_1}$? Yes because $y \in \mathbb{N_1}$, $G( \sqrt{\ln y})=e^{(\sqrt{\ln y})^2}$.
97562
Studying for an exam, a review question...  Given $G=\langle x,y|x^4=y^4=e,xyxy^{-1}=e\rangle$.    Show $|G|\leq16$.   For this, I want to consider that $x^3=x^{-1}$ and $y^3=y^{-1}$ based on our assumptions. I am a little lost as to how to put the second part, $xyxy^{-1}=e$ to show there are no more than 16 elements.   If $|G|=16$, find the center of the group and find a group that is isomorphic to $G/\langle y^2\rangle$.   I'm pretty sure the group is centerless.
102971
Let $GL(n,\mathbb{R})$ be the space of invertible $n \times n$ matrices over $\mathbb{R}$. I have already proved that $GL(n,\mathbb{R})$ is open in $\mathbb{R}^{n^2}$ and that it is not connected with the relative topology from $\mathbb{R}^{n^2}$. How do I prove that the map $f:GL(n,\mathbb{R})\rightarrow GL(n,\mathbb{R})$ given by $f(x):=x^{-1}$ is continuous?  Any help would be greatly apprecaited.   Additionally how do I prove that $g:GL(n,\mathbb{R})\times GL(n,\mathbb{R}) \rightarrow GL(n,\mathbb{R})$ given by $g(x,y):=xy$ (matrix multiplication) is continuous, where we take the relative topology from $\mathbb{R}^{n^2} \times \mathbb{R}^{n^2} \cong \mathbb{R}^{2n^2}$?
121202
$G(x) = ax + b$ where $a$ is not equal to $0$ and $b$ are real numbers.  Prove $G$ is a one-to-one correspondence.  I understand that for every $a$ there is a corresponding $b$-value that does not repeat; however, I do not understand how I can prove this.
108965
I'm trying to show that $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$. A hint would be nice.
27695
$F$ is an isomorphism from group $G$ onto group $B$. Prove $F^{-1}$ is an isomorphism from $B$ onto $G$.  I do not know which direction to go in.
95402
Let $G$ be a group and $c$ be a fixed element of $G$. Define a new operation $*$ on G by:   $$x*y=xdy$$   for all $x$ and $y$ in $G$, where $d$ is the inverse of $c$. Prove that $G$ is a group under the operation $*$.    If you want to check, this problem is in the book “A Course In Group Theory”, by John F. Humphreys (problem $3$, page $28$)   I know what are the axioms that $(G,*)$ must obey in order to be a group and I did show that all the axioms where obeyed except the inverse axiom.  My problem is that, whenever I try to verify what would the inverse be for some element $x$, I get an equation where the $x$ is cancelled in both sides.  Can please someone guide me?  Thank you!  (P.S.: This is NOT a homework assignment, I’m studying Group Theory by this book and got problems solving this question)
46530
Suppose $f$ and $g$ are continuous on $[a,b]$ and that $f(a) &lt; g(a)$ but $f(b) &gt; g(b)$.   Prove that $f(x) = g(x)$ for some $x \in [a,b]$.     I did try and actually did get as far as coming up with $f-g$. Would I set $h(x)= f(x)- g(x)$ and say $h(a) &lt;x&lt; h(b)$?
72531
Prove that $x^y+y^x&gt;1$ for all $(x, y)\in \mathbb{R_+^2}$.
69535
Let $f(x) = |x|$ for $x \in [-1,1]$ and extend $f$ to all of $\mathbb R$ by requiring $f(x+2)=f(x)$. Prove that $|f(x)-f(y)|\le|x-y|$ for all $x$ and $y$.   This is easy to observe on a graph but my proof is long and a bit tedious. Just curious if there is a quick/elegant way?
108703
I am trying to make a picture in my head so I can understand and remember the rules.  So if $f \circ g$ is onto, it is onto because the function $f$ maps every element from a set $B$ to a set $C$ (thus $f$ is onto) and if $f \circ g$ is one-to-one then every element from set $A$ is mapping an element of set $B$ (and thus is one-to-one).  If both $f$ and $g$ is onto then $f \circ g$ is onto and if both $f$ and $g$ is one-to-one then $f \circ g$ is one-to-one and if both $f$ and $g$ are bijective then $f \circ g$ is bijective?  If $f \circ g$ is bijective, we can't say anything, but that $f$ is onto and that $g$ is one-to-one?  If $f$ is onto and $g$ is one-to-one, nothing can be said?  If $g$ is one-to-one and $g$ is onto, nothing can be said?
97875
Exercise: If $f:X\rightarrow Y$ and $g: Y\rightarrow Z$ are both onto functions, prove or disprove, $g\circ f:X\rightarrow Z$ defined by $(g\circ f)(x)=g(f(x))$ is an onto function.  Just a little confused on how to do this problem. I know what an Onto function is, and what a composition function is but I don't know how to approach this problem, please help? thanks
79424
$f,g: \mathbb{R} \to \mathbb{R}$  I'd like to see a sketch for this proof.  [sorry for posting errors, I am on a cell phone]
64053
Possible Duplicate:                How to prove that isometry $f:X\to X$ is onto if $X$ is compact??  What about the case when $X$ is non-compact?  Any help would be appreciated.
87573
Given an undirected graph $G = (V,E)$ Prove the following a) Let $|V| = v$ and $|E|= e$. It holds that $2e &lt;= v^{2}-v$ b) If $|V| &gt; 3$ and $\exists! v_{0} \in V: d(v_{0})=1$, then there is a circuit in $G$.
95705
Let $f: X \rightarrow Y$ and $g: Y \rightarrow X$  If $f \circ g$ is surjective, then f is surjective, too.  I think that is true.   Question: How can I proove that? I have so far:   $\forall y \in Y: \exists x \in X : f(x)=y$  $\forall x \in Y : \exists y \in Y : f \circ g(y)= x$
15420
Let $f,g:[a,b]\subset \mathbb{R} \to \mathbb{R} $ be continuous functions with $f^2(x)=g^2(x)\neq 0$ for all $x\in[a,b]$. Show $f=g$ or $f=-g$ on $[a,b]$.