Quotient Spaces

0. The guiding question

A subspace $U\subseteq V$ is a collection of directions inside a vector space. Sometimes we want to ignore those directions.

For example, in $\mathbb R^3$, suppose $U$ is the horizontal $xy$-plane. If we decide that horizontal movement does not matter, then two points should be considered equivalent whenever they have the same height. What remains is a one-dimensional idea: the vertical coordinate.

A quotient space formalizes this idea.

The quotient $V/U$ is the vector space obtained from $V$ by declaring all vectors of $U$ to be zero.

The point is subtle: the elements of $V/U$ are not the original vectors of $V$. They are equivalence classes of vectors.

1. Equivalence modulo a subspace

Let $V$ be a vector space over a field $K$, and let $U\subseteq V$ be a subspace.

We define a relation on $V$ by

\[v\sim v'\]

\[v-v'\in U.\]

Equivalently, $v\sim v’$ if there exists some $u\in U$ such that

\[v=v'+u.\]

So $v$ and $v’$ are considered the same modulo $U$ if they differ only by a vector from $U$.

This relation is called equivalence modulo $U$, or congruence modulo $U$.

Theorem: $\sim$ is an equivalence relation

The relation $v\sim v’$ defined by $v-v’\in U$ is reflexive, symmetric, and transitive.

Proof idea

Reflexivity holds because

\[v-v=0\in U.\]

Symmetry holds because if $v-v’\in U$, then

\[v'-v=-(v-v')\in U.\]

Transitivity holds because if $v-v’\in U$ and $v’-v’‘\in U$, then

\[v-v''=(v-v')+(v'-v'')\in U.\]

All three steps use exactly the fact that $U$ is a subspace: it contains $0$, is closed under negatives, and is closed under addition.

2. Cosets: the elements of the quotient

The equivalence class of a vector $v\in V$ is

\[v+U=\{v+u\mid u\in U\}.\]

This is called a coset of $U$ in $V$.

Important:

$v+U$ is not one vector. It is a whole subset of $V$.

The quotient space $V/U$ is the set of all such cosets:

\[V/U=\{v+U\mid v\in V\}.\]

So an element of $V/U$ is a class of vectors that differ by elements of $U$.

The class of zero is

\[0+U=U.\]

Thus the zero vector of the quotient is the whole subspace $U$. This is the precise meaning of saying that $U$ has been collapsed to zero.

$Cosets of a subspace U (a line through the origin in R^2) form a family of parallel lines filling the whole space. Two vectors v and v' are equivalent modulo U exactly when they lie on the same parallel line, since their difference is then a vector inside U.$

3. Operations in the quotient

We define addition and scalar multiplication of cosets by

\[(v+U)+(w+U)=(v+w)+U,\]

and

\[\lambda(v+U)=(\lambda v)+U.\]

These definitions are natural: to add two classes, choose representatives, add them in $V$, and take the class of the result.

But there is a possible danger. A coset has many representatives. We must check that the result does not depend on which representatives we choose.

Theorem: the operations are well-defined

\[v+U=v'+U\]

and

\[w+U=w'+U,\]

then

\[(v+w)+U=(v'+w')+U.\]

Also, for every $\lambda\in K$,

\[(\lambda v)+U=(\lambda v')+U.\]

Proof idea

The equality $v+U=v’+U$ means $v-v’\in U$. Similarly, $w-w’\in U$. Therefore

\[(v+w)-(v'+w')=(v-v')+(w-w')\in U.\]

So $v+w$ and $v’+w’$ determine the same coset.

For scalar multiplication, if $v-v’\in U$, then

\[\lambda v-\lambda v'=\lambda(v-v')\in U.\]

Again, this uses exactly that $U$ is closed under addition and scalar multiplication.

Theorem: $V/U$ is a vector space

With the operations above, $V/U$ is a vector space over $K$.

Proof idea

Once the operations are well-defined, the vector space axioms follow from the corresponding axioms in $V$. For example, associativity of addition in $V/U$ follows because

\[((a+U)+(b+U))+(c+U) =((a+b)+c)+U\]

and

\[(a+U)+((b+U)+(c+U)) =(a+(b+c))+U,\]

which are equal because addition in $V$ is associative. The other axioms work the same way.

4. The quotient map

There is a natural map

\[\pi:V\to V/U\]

called the quotient map, defined by

\[\pi(v)=v+U.\]

It sends each vector to its equivalence class.

This map is linear:

\[\pi(v+w)=(v+w)+U=(v+U)+(w+U)=\pi(v)+\pi(w),\]

and

\[\pi(\lambda v)=(\lambda v)+U=\lambda(v+U)=\lambda\pi(v).\]

Its kernel is exactly $U$:

\[\ker\pi=U.\]

Indeed,

\[\pi(v)=0_{V/U}\]

means

\[v+U=U,\]

which happens exactly when $v\in U$.

This gives another useful interpretation:

The quotient map is the linear map that kills precisely the subspace $U$.

$The quotient map pi sends each vector in V to its coset in V/U. Each parallel coset on the left collapses to a single point on the right; the bold coset U itself collapses to the zero element of V/U, which is why ker pi = U.$

5. Geometric example: horizontal planes

Let

\[V=\mathbb R^3\]

and let

\[U=\{(x,y,0)\mid x,y\in\mathbb R\},\]

the $xy$-plane.

Two vectors $(x,y,z)$ and $(x’,y’,z’)$ are equivalent modulo $U$ when

\[(x,y,z)-(x',y',z')=(x-x',y-y',z-z')\]

lies in $U$. This happens exactly when

\[z=z'.\]

So the cosets are horizontal planes

\[z=c.\]

The quotient $\mathbb R^3/U$ is the space of horizontal layers. Each layer is determined by one number, its height $c$. Therefore

\[\mathbb R^3/U\cong \mathbb R.\]

This example captures the main intuition:

Quotienting by $U$ means that motion inside $U$ no longer matters.

Horizontal displacement is forgotten. Only vertical height remains.

$R^3 modulo the xy-plane: each coset is a horizontal layer z = c. Two points are equivalent exactly when they have the same height, so the quotient is parametrised by a single number z and is isomorphic to R.$

6. Example: quotienting a function space by invisible data

Let $X$ be a set, let $Y\subseteq X$, and let

\[V=\mathcal F(X,K)\]

be the vector space of all functions $X\to K$.

Define

\[I(Y)=\{g:X\to K\mid g(y)=0\text{ for all }y\in Y\}.\]

This is a subspace of $V$: it consists of all functions that vanish on $Y$.

Now consider the quotient

\[\mathcal F(X,K)/I(Y).\]

Two functions $f,f’\in\mathcal F(X,K)$ define the same coset if

\[f-f'\in I(Y),\]

which means

\[f(y)=f'(y)\]

for every $y\in Y$.

So the quotient forgets everything outside $Y$ and remembers only the values on $Y$.

There is a natural isomorphism

\[\mathcal F(X,K)/I(Y)\cong \mathcal F(Y,K)\]

given by

\[f+I(Y)\mapsto f|_Y.\]

Proof idea

The map is well-defined because if $f-f’$ vanishes on $Y$, then $f$ and $f’$ have the same restriction to $Y$. It is injective because if two functions have the same restriction to $Y$, then their difference vanishes on $Y$. It is surjective because any function on $Y$ can be extended to some function on $X$. Linearity is inherited from pointwise operations.

This example is conceptually important: quotienting is not only a geometric operation. It is a way of keeping some information and declaring other information irrelevant.

7. Dimension of a quotient

Now assume $V$ is finite-dimensional and $U\subseteq V$.

Theorem: dimension formula for quotients

\[\dim(V/U)=\dim V-\dim U.\]

Proof idea

Choose a basis of $U$, say

\[(e_1,\dots,e_m),\]

and extend it to a basis of $V$:

\[(e_1,\dots,e_m,e_{m+1},\dots,e_n).\]

The vectors $e_1, \dots,e_m$ disappear in the quotient because they lie in $U$:

\[e_i+U=U=0_{V/U}\]

for $1\leq i\leq m$.

The surviving classes

\[e_{m+1}+U, \dots, e_n+U\]

form a basis of $V/U$.

They span because every vector $v\in V$ can be written as

\[v=a_1e_1+\cdots+a_me_m+a_{m+1}e_{m+1}+\cdots+a_ne_n,\]

and in the quotient the first $m$ terms vanish:

\[v+U=a_{m+1}(e_{m+1}+U)+\cdots+a_n(e_n+U).\]

They are linearly independent because a relation

\[a_{m+1}(e_{m+1}+U)+\cdots+a_n(e_n+U)=U\]

means

\[a_{m+1}e_{m+1}+\cdots+a_ne_n\in U.\]

But the only linear combination of $e_{m+1},\dots,e_n$ that lies in $U=\langle e_1,\dots,e_m\rangle$ is the zero combination, because the full list is a basis of $V$. Thus all coefficients are zero.

So $V/U$ has $n-m$ basis vectors, and

\[\dim(V/U)=n-m=\dim V-\dim U.\]

8. Codimension

The codimension of $U$ in $V$ is defined by

\[\operatorname{codim}_V U=\dim(V/U).\]

If $V$ is finite-dimensional, then

\[\operatorname{codim}_V U=\dim V-\dim U.\]

The dimension of $U$ counts how many independent directions lie inside $U$. The codimension counts how many independent directions remain after $U$ has been collapsed.

Examples:

\[\operatorname{codim}_{\mathbb R^3}(\text{plane through }0)=1.\]

A plane in $\mathbb R^3$ leaves one independent transverse direction.

\[\operatorname{codim}_{\mathbb R^3}(\text{line through }0)=2.\]

A line in $\mathbb R^3$ leaves two independent transverse directions.

\[\operatorname{codim}_V V=0.\]

If we collapse all of $V$, nothing remains.

\[\operatorname{codim}_V \{0\}=\dim V.\]

If we collapse only zero, the whole space remains.

9. Quotients and complements

A quotient space is abstract: its elements are cosets. But sometimes we can choose a concrete representative from each coset.

Suppose $W\subseteq V$ is a subspace such that

\[V=U\oplus W.\]

This means every vector $v\in V$ can be written uniquely as

\[v=u+w,\]

where $u\in U$ and $w\in W$.

Then $W$ gives one representative of each quotient class.

$Direct sum decomposition V = U + W: the line U and a transverse line W meet only at zero. Every vector v in the plane splits uniquely as v = u + w with u on U and w on W, and the point where W crosses the coset of U through v is exactly w.$

Theorem: a complement represents the quotient

\[V=U\oplus W,\]

then

\[V/U\cong W.\]

More precisely, the map

\[\varphi:W\to V/U, \qquad \varphi(w)=w+U\]

is an isomorphism.

Proof idea

The map is linear because quotient operations are defined using representatives:

\[\varphi(w_1+w_2)=(w_1+w_2)+U=(w_1+U)+(w_2+U).\]

It is injective because if

\[w+U=w'+U,\]

then $w-w’\in U$. But $w-w’\in W$ too. Since $V=U\oplus W$, we have

\[U\cap W=\{0\}.\]

Thus $w-w’=0$, so $w=w’$.

It is surjective because every class $v+U$ has a representative in $W$. Indeed, write

\[v=u+w.\]

Then

\[v+U=(u+w)+U=w+U.\]

The $U$-part disappears in the quotient.

So each coset meets $W$ in exactly one point.

10. Quotient versus complement

The theorem above does not say that $V/U$ is literally equal to $W$.

They are different kinds of objects:

$W$ consists of vectors of $V$;
$V/U$ consists of cosets of $U$.

The theorem says that once a complement $W$ has been chosen, $W$ gives a concrete model of the quotient.

The difference matters because complements are not unique.

For example, in $\mathbb R^2$, let

\[U=\langle (1,0)\rangle,\]

the $x$-axis. The $y$-axis is a complement, but so is the line

\[\langle(1,1)\rangle.\]

In fact, every line through the origin except the $x$-axis is a complement of $U$.

$With U the x-axis in R^2, any other line through the origin is a complement: W_1 (the y-axis), W_2, W_3 are three different choices. Each intersects every horizontal coset of U in exactly one point and is therefore isomorphic to the quotient V/U, even though the three lines are not equal as subspaces.$

Each complement is isomorphic to $\mathbb R^2/U$, but no one of them is forced by $U$ alone.

The quotient $V/U$, however, is canonical: it is determined only by $V$ and $U$.

So the correct mental picture is:

\[\boxed{\text{A quotient is canonical.}}\] \[\boxed{\text{A complement is a choice of representatives.}}\]

11. Common confusions

Confusion 1: $V/U$ consists of vectors outside $U$

No. The quotient consists of cosets $v+U$, not individual vectors. A vector $v$ is only a representative of the coset $v+U$.

Confusion 2: the zero of $V/U$ is the vector $0$

Not exactly. The zero element of $V/U$ is the coset

\[0+U=U.\]

So the whole subspace $U$ becomes zero.

Confusion 3: quotienting removes vectors

It is better to say that quotienting identifies vectors. Vectors that differ by an element of $U$ become the same element of $V/U$.

Confusion 4: a complement is the same thing as the quotient

A complement $W$ is a concrete subspace inside $V$. The quotient $V/U$ is a space of cosets. If $V=U\oplus W$, then $W\cong V/U$, but the isomorphism depends on the choice of $W$.

12. Summary

The quotient $V/U$ is built by declaring two vectors equivalent if they differ by an element of $U$:

\[v\sim v' \quad\Longleftrightarrow\quad v-v'\in U.\]

Its elements are cosets

\[v+U.\]

The operations are

\[(v+U)+(w+U)=(v+w)+U,\]

and

\[\lambda(v+U)=(\lambda v)+U.\]

The zero element is

\[U.\]

If $V$ is finite-dimensional, then

\[\dim(V/U)=\dim V-\dim U.\]

This number is the codimension of $U$ in $V$.

If $W$ is a complement of $U$, meaning

\[V=U\oplus W,\]

then

\[V/U\cong W.\]

The quotient is the abstract space of directions left after ignoring $U$. A complement is a concrete choice of one representative for each quotient class.