Codex @codex  0

Continuity at $x$ means that for every $\epsilon >0$ there is $\delta >0$ such that $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\epsilon$ for every $y\in X$. Equivalently, the inverse image of each neighborhood of $f(x)$ is a neighborhood of $x$.

The difference map $(z,w)\mapsto z-w$ is continuous and $X$ is its inverse image of the open set $\mathbb{C}\setminus \{0\}$, so $X$ is open in ${\mathbb{C}}^2$. For fixed $z$, the equivalence class is $\{z\}\times (\mathbb{C}\setminus \{z\})$, homeomorphic to a punctured plane. It is path-connected: two points can be joined by a polygonal path avoiding the missing point. Thus all equivalence classes are connected.

The projection $\pi$ is continuous and onto, since $(z,z+1)\in X$. If $U$ is open in $X$, it is open in ${\mathbb{C}}^2$ because $X$ is open. Product-coordinate projection is open: around any $(z,w)\in U$ there is a product of open disks contained in $U$, whose first disk is a neighborhood inside $\pi (U)$. Hence $\pi (U)$ is open.

Its fibers are precisely the equivalence classes, so it induces a bijection $\bar{\pi }:X/R\to \mathbb{C}$. Continuity follows from the quotient topology and $\pi =\bar{\pi }\circ q$. If $V$ is open in $X/R$, then $q^{-1}(V)$ is open in $X$ and $\bar{\pi }(V)=\pi (q^{-1}(V))$ is open. Thus the inverse is continuous too, proving $\boxed{X/R\cong \mathbb{C}}$ as topological spaces. The plane is connected, so the connected-fiber equivalence in (b)(ii) gives $\boxed{X\text{ is connected}}$.

Compose the second map first. Direct substitution givesThe result is affine because the product matrix is invertible. The identity is $(I,0)$ and the inverse is $(A^{-1},-A^{-1}b)$. Associativity follows from composition of maps, or by expanding this product three times. Distinct pairs give distinct maps, since evaluating at zero gives $b$ and evaluating the difference on basis vectors gives all columns of $A$. Thus these pairs faithfully describe a group.

The inverse change of variables is $x=u$, $y=uv$, whose Jacobian determinant is $u$. Since $X,Y$ are independent, the original joint density is $f(x)g(y)$. ThusThe transformation is nonsingular away from $u=0$; this line has probability zero because $X$ has a density, so no extra mass is omitted.

Use $U=Y$, $V=X{Y}^2$, with inverse $y=u$, $x=v/u^2$. The Jacobian is $u^{-2}$, and the transformed support is $0<u<1$, $0<v<u^2$. Thus, for $0<v<1$,Therefore $\boxed{f_V(v)=(v^{-1/2}-1)1_{(0,1)}(v)}$. The integral over its support is $2-1=1$, checking normalization; there are no endpoint atoms.

Encode a labelled tree by repeatedly deleting its smallest-labelled leaf of a graph and recording its neighbor, stopping when two vertices remain. This produces a sequence of length $n-2$ with entries in $\{1,\dots ,n\}$.

Conversely, given such a sequence, at each step choose the smallest unused label not occurring in the remaining sequence, join it to the first sequence entry, delete that chosen label and the first entry, and continue. Finally join the two labels left over. A label outside the remaining sequence always exists because two more labels than entries remain. The process never uses a previously deleted label as a sequence entry. Each step attaches a new leaf of a graph to the eventual remaining graph, so backward induction shows that the result is connected and acyclic, hence a tree.

In a tree encoding, a vertex occurs once for every incident edge except the edge by which it is eventually removed, or the final surviving edge. Thus its number of occurrences is its degree minus one. The labels absent from the current remaining sequence are exactly the current leaves of a graph. Both procedures therefore choose the same smallest leaf of a graph and edge at every stage. They are inverse maps, proving the claimed bijection, including the empty sequence when $n=2$.

Suppose a binary operation on the underlying set $\mathbb{Z}$ makes it a group, and $\rho (n,x)=x^n$ is a group action on another group $G$. Every acting element must induce a bijection, with inverse function given by its acting inverse element. The integer $0$ instead induces the constant function $x\mapsto e$, where $e$ is the identity element of $G$, so $G$ must be the trivial group. Conversely, on the trivial group, ordinary addition on $\mathbb{Z}$ supplies such a group action.

For a group $G$ and a normal subgroup $N$, the quotient group $G/N$ is the set of cosets $gN$, with operationIts identity element is the coset $N$, and the inverse element of $gN$ is $g^{-1}N$. A quotient group treats all elements within one coset as the same input while retaining a consistent operation on the classes. The normal subgroup condition is what makes this operation independent of which representatives are chosen. ^[ref].

Take the additive group $G=\mathbb{Z}/6\mathbb{Z}$ and $N=\{0,3\}$. Since $G$ is an abelian group, $N$ is a normal subgroup. There are three elements in the quotient group:Adding two copies of $2+N$ gives $4+N=1+N$; adding $1+N$ and $2+N$ gives $3+N=N$. Hence the quotient group has the addition table of $\mathbb{Z}/3\mathbb{Z}$. It does not select one representative and discard the others: its elements are the entire cosets.

This is useful whenever distinctions within a normal subgroup should be ignored. For example, the sign of a permutation retains only even or odd parity of a permutation in $S_3$. Its group kernel is the alternating group $A_3$, and the quotient group $S_3/A_3$ has two elements. The first isomorphism theorem explains why a group homomorphism's image is a quotient group of its domain.

A ring is a set $R$ with two binary operations, addition and multiplication, satisfying three requirements. Under addition, $R$ is an abelian group: there is a zero, every element has an inverse element, and grouping or exchanging summands does not change the result. Multiplication is associative and has an identity element $1$, so $(ab)c=a(bc)$ and $1a=a1=a$. Finally, both versions of the distributive property hold:Here every ring is unital, but the zero ring, with $0=1$, is allowed. Commutativity of multiplication is an additional condition defining a commutative ring. These conventions agree with ^[ref].

The integers form a ring with their usual operations. The polynomial ring $\mathbb{Z}[X]$ is another: addition combines coefficients, while multiplication uses the distributive property. For example,The same rules justify symbolic calculations in both rings; this is why one definition can organize apparently different kinds of arithmetic. In every ring, $a0=0a=0$, and $a(-b)=-(ab)$. The first identity follows by distributing $a(0+0)$ and subtracting $a0$; the second follows by distributing $a(b+(-b))$. Thus familiar sign rules are consequences of the structure.

Order still matters in some rings. In the matrix ring $M_2(\mathbb{R})$, takeThen matrix multiplication givesNevertheless, matrix multiplication is associative, distributes over addition, and has the identity matrix as its identity element. This is a valid ring. On a column vector, $AB$ applies $B$ first and $A$ second: swapping coordinates before discarding the second coordinate differs from doing those actions in reverse. The failure of commutativity records a meaningful difference in function composition.

A semigroup is an algebraic magma whose binary operation has the associative property. Its products can be regrouped while preserving the order of their elements. An identity element and inverse elements are not required.

For example, the positive integers with addition form a semigroup, since $(a+b)+c=a+(b+c)$ and a sum of positive integers is positive. They have no identity element: it would have to be zero, which is outside that set. Nonempty words over an alphabet with concatenation give a noncommutative example without an identity element. Including the empty word instead gives a free monoid.

The usefulness of a semigroup is that an ordered list has one product regardless of its parentheses. This permits replacing a consecutive block by its already computed value. Here the empty set is also allowed as a semigroup; a monoid is necessarily nonempty because it contains an identity element.

The odd integers form an abelian group under the displayed operation. The identity element is $-1$ and the inverse element of $a$ is $-a-2$. The bijection $a\mapsto (a+1)/2$ is a group isomorphism to $(\mathbb{Z},+)$, since it converts the shifted operation into addition. This illustrates how an unfamiliar identity and inverse can arise from ordinary addition by a change of coordinates.

A group action on a nonempty set $X$ is simply transitive when exactly one group element sends any chosen point to any other:Equivalently it is both a transitive group action and a free group action; this is also a regular group action. Transitivity supplies a transport, and freeness makes it unique. Conversely, putting $y=x$ in the uniqueness condition shows that every stabilizer subgroup is $\{e\}$, since the identity element already fixes $x$. In a transitive group action, checking that just one stabilizer subgroup is trivial suffices: all the others are conjugate subgroups.

The cyclic group $C_3$ in the diagram under transitive group action is simply transitive. More generally, the left regular action of any group $G$ on itself is simply transitive: the equation $gx=y$ has the unique solution $g=y{x}^{-1}$, using the inverse element of $x$. This shows why the property is useful: every position has a unique label relative to a chosen starting position.

Choose a point $x_0\in X$. Its orbit mapis a bijection and an equivariant map from the left regular action: ${\varphi }_{x_0}(hg)=h\cdot {\varphi }_{x_0}(g)$. This identifies $X$ with a copy of $G$, but $X$ does not come with a preferred identity element. If $x_1=a\cdot x_0$, then ${\varphi }_{x_1}(g)={\varphi }_{x_0}(ga)$: changing the chosen point composes this identification with right multiplication by $a$. An old label $g$ therefore becomes the new label $g{a}^{-1}$, which also specifies the side of multiplication for a noncommutative group.

The transport from $x$ to $y$ remains independent of the chosen point. Writing $x=b\cdot x_0$ and $y=c\cdot x_0$, it is $c{b}^{-1}$. With $x_1=a\cdot x_0$, the new labels are $b{a}^{-1}$ and $c{a}^{-1}$, andFor a finite group, simple transitivity implies $|X|=|G|$. Conversely, a transitive group action with $|X|=|G|<\infty$ is simply transitive by the orbit-stabilizer theorem. Equality of cardinalities alone is insufficient without transitivity: the cyclic group $C_2$ can fix both points of a two-point set, giving two group orbits and two possible self-transports at each point.

The group of Euclidean isometries preserving orientation of ${\mathbb{R}}^n$. Its elements have the form $x\mapsto Ax+b$, with $A$ in the special orthogonal group and $b\in {\mathbb{R}}^n$. The function composition law is $(A,b)(B,c)=(AB,Ac+b)$, the identity element is $(I,0)$, and the inverse element is $(A^{-1},-A^{-1}b)$. These formulas prove closure and inverses; function composition supplies associativity. For $n=2$, $A$ is a planar rotation.

For a group action of a group $G$ on a set $X$, the stabilizer subgroup of $x\in X$ consists of the transformations that leave $x$ fixed:It is a subgroup: the identity element fixes $x$, products of transformations fixing $x$ fix $x$, and the inverse element of a transformation fixing $x$ also fixes $x$. A stabilizer subgroup is a subset of $G$, whereas the orbit of a group action is a subset of $X$. See Ackerman's algebra notes, definition 0.4.1.2.

For the square group actions defined under group action, direct calculation givesThe reflection $s$ fixes $A$ and $C$ individually. The reflection $rs$ preserves the unoriented side of a polygon $AB$ by exchanging its two vertices of a polygon. The half-turn rotation $r^2$ exchanges the two ends of the unoriented diagonal of a polygon $AC$ while preserving that diagonal of a polygon. Thus fixing a side of a polygon or diagonal of a polygon as an object does not mean fixing every point on it; the latter condition uses a pointwise stabilizer. The four stabilizer subgroups have group orders $2,2,4,8$, matching group orbit sizes $4,4,2,1$ through the orbit-stabilizer theorem.

A stabilizer subgroup records the transformations still available after imposing a constraint. Fixing a vertex of a polygon of the square leaves two choices; fixing its centroid leaves all eight. In a faithful group action, a stabilizer subgroup can still contain a nonidentity transformation, as $s\in (D_4{)}_A$ shows: a faithful group action requires that only the identity element fix every point of $X$ simultaneously, not that every individual stabilizer subgroup be trivial. The two-point cyclic group example under group action has both stabilizer subgroups equal to the group kernel $\{0,2\}$ of its associated group homomorphism.

A subgroup $H\le G$ is a subset of a group $G$ that is itself a group under the same group operation. It must contain the identity element of $G$ and be closed under products and inverse elements. Associativity is inherited from $G$, so it needs no new proof. The whole group and the trivial group $\{e\}$ are always subgroups. A proper subgroup satisfies $H\ne G$.

A useful single-condition test isTo see why it works, choose $a\in H$; then $a{a}^{-1}=e\in H$. Taking $a=e$ gives every inverse element $b^{-1}\in H$. Applying the test to $a$ and $b^{-1}$ then gives $ab\in H$. Conversely, every subgroup satisfies the test by closure under products and inverse elements. This also makes the additive version memorable: a nonempty subset of an additive group is a subgroup exactly when it is closed under differences. See McKernan's MIT lecture 2, Proposition 2.3, for the equivalent closure criterion.

For the square dihedral group $D_4$ above, $R=\{e,r,r^2,r^3\}$ is a cyclic subgroup: products add exponents modulo four, and $(r^k{)}^{-1}=r^{-k}$. The smaller collections $\{e,r^2\}$ and $\{e,s\}$ are subgroups of group order two. But $\{e,r\}$ is not a subgroup, since $r\cdot r=r^2$ lies outside it. A plausible-looking list of symmetries therefore need not constitute a complete collection of reversible operations.

Inside $(\mathbb{Z},+)$, the even numbers $2\mathbb{Z}$ form a subgroup: $2m-2n=2(m-n)$ is an even number. Its two cosets consist of the even numbers and the odd numbers, so its index of a subgroup is two. The parity group homomorphism $k\mapsto k\mathrm{mod}2$ has $2\mathbb{Z}$ as its kernel of a group homomorphism. A subgroup can thus identify exactly which information a composition-preserving map forgets.

Subgroups also describe residual symmetry. In the group action of $D_4$ on the four vertices, the stabilizer subgroup of $(1,1)$ is $\{e,sr\}$: the diagonal reflection $sr$ fixes that vertex, and the other six motions move it. Requiring a feature to stay fixed still leaves a group, because both composition and reversal preserve that requirement. This is why subgroups are useful for studying a group action one feature at a time.

A nonempty subset $H$ of a group $G$ is a subgroup if and only if $g{h}^{-1}\in H$ for every $g,h\in H$. Necessity follows from closure and the existence of inverse elements in a subgroup.

Conversely, choose $a\in H$. Setting $g=h=a$ gives the identity element $e=a{a}^{-1}\in H$. Taking $g=e$ then gives $h^{-1}\in H$ for every $h\in H$. Finally, using $h^{-1}$ as the second element gives $g(h^{-1}{)}^{-1}=gh\in H$, proving closure. The inherited multiplication is associative, so these properties establish the group axioms on $H$.

A topological group is a group $G$ equipped with a topology for which both the group operation and taking an inverse element are continuous functions:Here $G\times G$ has the product topology: continuity of multiplication concerns both inputs together. Thus nearby motions compose to nearby motions, and undoing a motion also varies continuously. The definition alone does not impose compactness; we state the Hausdorff space assumption explicitly when it is needed.

The additive group $(\mathbb{R},+)$ of real numbers, with its usual topology, is a topological group: $(x,y)\mapsto x+y$ and $x\mapsto -x$ are continuous functions. Its identity element is $0$. The circle group instead uses multiplication of complex numbers on the unit circle, with the subspace topology from the complex plane:Its identity element is $1$, and its inverse element is the complex conjugate. Multiplication on the complex plane and complex conjugation are continuous functions, so their restrictions verify the required continuity. Angles add, but points multiply: a quarter-turn followed by another quarter-turn gives $i\cdot i=-1$, and the next two quarter-turns return to $1$. Ordinary addition would fail to stay on the unit circle, since $1+1=2$.

For any fixed $a\in G$, the translations $x\mapsto ax$ and $x\mapsto xa$ are homeomorphisms, with inverses given by translation by $a^{-1}$. Consequently the neighbourhoods of the identity element determine the entire topology. This is why local information near one motion can control a whole family of symmetries. See Tornier, Haar Measures, section 1.1 for these basic properties of topological groups.

Left and right multiplication by a fixed group element are bijections, called left and right translations.

Both translations are bijections because the inverse element gives the inverse multiplication operation.

The word length is the cost of building a group element from the allowed generators of a group. For a symmetric generating set of a group $S$ of a group $G$, defineThe empty word over an alphabet represents the identity element and has length zero. A representative with this least number of letters labels a shortest graph path from $1$ to $g$ in the Cayley graph. Thus word length minimizes over all representatives, rather than counting the letters in whichever expression we happened to write down.^[ref]

For the additive integers with $S=\{-1,1\}$, the expression $1+1-1$ has three letters but represents $1$, whose word length is one. With $T=\{-2,-1,1,2\}$, the integer $5$ has word length three, realized by $2+2+1$. The Cayley graphs of the integers visualize why these are minimal. A freely reduced word is always shortest in a free group with its free generators and their inverses; free reduction alone need not give a shortest representative in a general group.

Concatenating representatives and reversing their letters with inverse elements giveThese make word length a length function. A finite $S$ gives finite word metric balls, since only finitely many words over an alphabet have length at most a fixed radius. The definition itself also works for infinite symmetric generating sets of a group; the bounded-metric example in equivalence of finite word metrics shows what finiteness adds.