A [vector space](/page/Vector%20Space) often contains information we want to treat as invisible. In geometry, two vectors may differ by a direction we have decided to collapse. In linear algebra, two solutions of an equation may differ by an element of the kernel and so produce the same output. In algebra, two elements of a module may be indistinguishable modulo a submodule. The quotient space is the construction that turns this deliberate loss of information into a new space rather than a vague instruction to ignore something.

The danger is that ignoring a subspace by informal language does not automatically produce algebra. If $v$ and $w$ are to count as the same whenever $v-w$ lies in a chosen subspace $U$, then addition and scalar multiplication must not depend on which representatives are chosen. The quotient construction packages exactly the [equivalence relation](/page/Equivalence%20Relation) for which these operations are well-defined.

[example: Lines Parallel to a Plane] Let $V=\mathbb R^3$ and let $U=\operatorname{span}(e_1,e_2)=\{(a,b,0):a,b\in\mathbb R\}$ be the $xy$-plane. We show that two vectors represent the same coset modulo $U$ exactly when they have the same height coordinate. For every $(x,y,z)\in\mathbb R^3$, the displacement from $(0,0,z)$ to $(x,y,z)$ is horizontal: \begin{align*} (x,y,z)-(0,0,z)=(x,y,0)=xe_1+ye_2\in U. \end{align*} Conversely, if $(a,b,c)\in (x,y,z)+U$, then for some $s,t\in\mathbb R$, \begin{align*} (a,b,c)=(x,y,z)+(s,t,0)=(x+s,y+t,z). \end{align*} So every point of $(x,y,z)+U$ has third coordinate $z$, and every point $(r,s,z)$ with third coordinate $z$ lies in that coset because \begin{align*} (r,s,z)=(x,y,z)+(r-x,s-y,0) \end{align*} with $(r-x,s-y,0)\in U$. Hence \begin{align*} (x,y,z)+U=\{(r,s,z):r,s\in\mathbb R\}=(0,0,z)+U. \end{align*} Now define $\pi:\mathbb R^3\to\mathbb R$ by $\pi(x,y,z)=z$. Its kernel is exactly $U$, since \begin{align*} \ker(\pi)=\{(x,y,z):z=0\}=\{(x,y,0):x,y\in\mathbb R\}=U. \end{align*} For each $h\in\mathbb R$, the fibre over $h$ is \begin{align*} \pi^{-1}(\{h\})=\{(x,y,z):z=h\}=\{(x,y,h):x,y\in\mathbb R\}=(0,0,h)+U. \end{align*} Thus the cosets of $U$ are exactly the horizontal planes, and the quotient $\mathbb R^3/U$ records only the common height of all vectors in such a plane. [/example]

The example gives the central picture: a quotient is not a subset of the original space. It is a new object whose elements are entire parallel copies of the subspace being collapsed. This distinction prevents a common mistake: a quotient space is not obtained by deleting $U$ from $V$, but by declaring every translate $v+U$ to be a single point.

## Definition

The central construction begins with a decision: two vectors should count as the same when their difference lies in the subspace being ignored. The quotient is the space of these equivalence classes, equipped with operations inherited from the original vector space.

[definition: Quotient Space] Let $V$ be a vector space over a field $k$, and let $U\subset V$ be a linear subspace. For $v\in V$, write \begin{align*} v+U:=\{v+u:u\in U\}. \end{align*} The quotient space of $V$ by $U$ is the set \begin{align*} V/U:=\{v+U:v\in V\}, \end{align*} equipped with the operations \begin{align*} +:(V/U)\times(V/U)\to V/U,\qquad (v+U,w+U)\mapsto (v+w)+U, \end{align*} and \begin{align*} \cdot:k\times(V/U)\to V/U,\qquad (\lambda,v+U)\mapsto (\lambda v)+U. \end{align*} [/definition]

The definition deliberately names the operations before proving that they make sense. That is the main technical point: a quotient space is only legitimate if changing representatives does not change the resulting class. The next theorem is therefore not a routine afterthought; it is the result that turns the set of cosets into a usable vector space.

[quotetheorem:9489]

This theorem is the algebraic license to compute with representatives. Once the quotient has been formed, cosets may be added by adding any representatives, because all choices produce the same coset.

## Representatives and Well-Defined Operations

The quotient is easiest to use when we understand when two names describe the same new point. The notation $v+U$ is not meant to remember $v$; it remembers the whole translate of $U$ through $v$.

[definition: Coset of a Subspace] Let $V$ be a vector space over a field $k$, and let $U\subset V$ be a linear subspace. For $v\in V$, the coset of $U$ represented by $v$ is the subset \begin{align*} v+U:=\{v+u:u\in U\}\subset V. \end{align*} [/definition]

A coset is the algebraic version of a parallel affine copy of $U$. If $U$ is a plane through the origin in $\mathbb R^3$, its cosets are planes parallel to it. The representative $v$ is not part of the data of the coset; many different vectors can name the same translate. This creates the first practical problem in any quotient calculation: expressions such as $v+U=w+U$ are statements about two subsets of $V$, but computations usually start with the representatives $v$ and $w$. We need a representative-level criterion for equality, otherwise every calculation in $V/U$ would require comparing two entire translates.

The obstruction is that equality of subsets is too large a test for routine algebra. What is needed is an intrinsic condition on the two representatives: exactly how far apart may $v$ and $w$ be while still naming the same collapsed point? In a quotient, the permitted error should be precisely a vector of $U$.

[remark: Representative Criterion for Subspace Cosets] For $v,w\in V$, the cosets $v+U$ and $w+U$ are equal exactly when their difference lies in the collapsed subspace: \begin{align*} v+U=w+U \quad\Longleftrightarrow\quad v-w\in U. \end{align*} Thus changing a representative by an element of $U$ does not change the coset, and every equality of cosets comes from such a change of representative. [/remark]

This is the moment where the quotient stops being mysterious. Equality in $V/U$ means equality after forgetting all directions in $U$.

[example: Quotient by a Line in the Plane] Let $V=\mathbb R^2$ and let $U=\operatorname{span}(1,1)=\{(t,t):t\in\mathbb R\}$. For $(a,b)\in\mathbb R^2$, its coset is \begin{align*} (a,b)+U=\{(a,b)+(t,t):t\in\mathbb R\}=\{(a+t,b+t):t\in\mathbb R\}. \end{align*} Thus each coset is a line parallel to the line spanned by $(1,1)$. We now compute exactly when two representatives give the same coset. If $(a,b)-(c,d)\in U$, then there is $t\in\mathbb R$ such that \begin{align*} (a-c,b-d)=t(1,1)=(t,t). \end{align*} So $a-c=t$ and $b-d=t$, hence \begin{align*} a-b=(c+t)-(d+t)=c-d. \end{align*} Conversely, if $a-b=c-d$, then $a-c=b-d$. Setting $t=a-c=b-d$ gives \begin{align*} (a,b)-(c,d)=(a-c,b-d)=(t,t)=t(1,1)\in U. \end{align*} Therefore $(a,b)$ and $(c,d)$ determine the same coset exactly when $a-b=c-d$. Define $T:\mathbb R^2\to\mathbb R$ by $T(a,b)=a-b$. The calculation above shows that $T$ is constant on each coset, so the rule \begin{align*} \widetilde T((a,b)+U)=a-b \end{align*} is well-defined on $\mathbb R^2/U$. Its kernel is trivial in the quotient: if $\widetilde T((a,b)+U)=0$, then $a-b=0$, so $a=b$ and \begin{align*} (a,b)=a(1,1)\in U. \end{align*} Hence $(a,b)+U=U$, the zero coset. The map is also surjective, since for every $r\in\mathbb R$, \begin{align*} \widetilde T((r,0)+U)=r-0=r. \end{align*} Thus $\widetilde T:\mathbb R^2/U\to\mathbb R$ is a linear isomorphism. The quotient remembers the transverse quantity $a-b$ and forgets motion in the direction $(1,1)$. [/example]

Not every formula on representatives descends to the quotient. A formula is allowed only if it gives the same answer on every representative of the same coset; otherwise it is still a formula on $V$, not a function on $V/U$.

[example: A Formula That Does Not Descend] Let $V=\mathbb R^2$ and let $U=\operatorname{span}(1,1)=\{t(1,1):t\in\mathbb R\}$. Define $F:\mathbb R^2\to\mathbb R$ by $F(a,b)=a$. The vectors $(0,0)$ and $(1,1)$ determine the same coset modulo $U$, because \begin{align*} (1,1)-(0,0)=(1,1)=1(1,1)\in U. \end{align*} Equivalently, $(1,1)\in U$, so \begin{align*} (1,1)+U=U=(0,0)+U. \end{align*} However, the proposed value of the rule $(a,b)+U\mapsto a$ depends on which representative of this coset is used: \begin{align*} F(0,0)=0 \end{align*} and \begin{align*} F(1,1)=1. \end{align*} Thus the single coset $U$ would have to be assigned both $0$ and $1$, so the rule $(a,b)+U\mapsto a$ does not define a function $\mathbb R^2/U\to\mathbb R$. By contrast, replacing $(a,b)$ by another representative in the same coset means replacing it by \begin{align*} (a,b)+t(1,1)=(a+t,b+t) \end{align*} for some $t\in\mathbb R$. The quantity $a-b$ is unchanged under this replacement, since \begin{align*} (a+t)-(b+t)=a+t-b-t=a-b. \end{align*} So $(a,b)+U\mapsto a-b$ is well-defined, while $(a,b)+U\mapsto a$ still remembers information in the collapsed direction. [/example]