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Quotient spaces 1 / 10

Quotient spaces

2014-08-29 09:00

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Quotient spaces j Denition 2 / 10

1 Denition

2 Maps going out of quotients

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Quotient spaces j Denition 2 / 10

Quotient maps

Denition (Quotient map)Let X; Y be topological spaces and p: X! Y a surjectivemap. Then p is called a quotient map if U Y is open ifand only if p 1(U) is open in X.

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Quotient spaces j Denition 2 / 10

Quotient maps

Denition (Quotient map)Let X; Y be topological spaces and p: X! Y a surjectivemap. Then p is called a quotient map if U Y is open ifand only if p 1(U) is open in X.

Remarks

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Quotient spaces j Denition 2 / 10

Quotient maps

Denition (Quotient map)

Let X; Y be topological spaces and p: X! Y a surjectivemap. Then p is called a quotient map if U Y is open ifand only if p 1(U) is open in X.

Remarks

p is a quotient map if and only if p is surjective and

p 1(A) is closed in X if and only if A Y closed.

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Quotient spaces j Denition 2 / 10

Quotient maps

Denition (Quotient map)

Let X; Y be topological spaces and p: X! Y a surjectivemap. Then p is called a quotient map if U Y is open ifand only if p 1(U) is open in X.

Remarks

p is a quotient map if and only if p is surjective and

p 1(A) is closed in X if and only if A Y closed.

Maps which are open, continuous and surjective arequotient maps.

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Quotient spaces j Denition 2 / 10

Quotient maps

Denition (Quotient map)

Let X; Y be topological spaces and p: X! Y a surjectivemap. Then p is called a quotient map if U Y is open ifand only if p 1(U) is open in X.

Remarks

p is a quotient map if and only if p is surjective and

p 1(A) is closed in X if and only if A Y closed.

Maps which are open, continuous and surjective arequotient maps.

Maps which are closed, continuous and surjective are

quotient maps.

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Quotient spaces j Denition 3 / 10

Saturated subsets

Denition (Fiber)Let p: X! Y be a surjective map between sets andy2 Y. The inverse image set

p`1(fyg) = fx2 X j p(x) = yg

will be denoted as p 1(y) and referred to as the ber of y.

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Quotient spaces j Denition 3 / 10

Saturated subsets

Denition (Fiber)Let p: X! Y be a surjective map between sets andy2 Y. The inverse image set

p`1(fyg) = fx2 X j p(x) = yg

will be denoted as p 1(y) and referred to as the ber of y.

Denition (Saturated set)

Let p: X! Y be a surjective map between sets. We saythat A X is saturated (with respect to p) ifA \ p 1(y) 6= ; implies p 1(y) A

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Quotient spaces j Denition 4 / 10

Remarks

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Quotient spaces j Denition 4 / 10

Remarks

A is saturated if and only if A = p 1(p(A)).

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Quotient spaces j Denition 4 / 10

Remarks

A is saturated if and only if A = p 1(p(A)).

If U Y, then p 1(U) is saturated.

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Q o e p e j e o /

Remarks

A is saturated if and only if A = p 1(p(A)).

If U Y, then p 1(U) is saturated.

p is a quotient map if and only if p is surjective,continuous, and A saturated and open implies p(A) open.

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Q j /

Remarks

A is saturated if and only if A = p 1(p(A)).

If U Y, then p 1(U) is saturated.

p is a quotient map if and only if p is surjective,continuous, and A saturated and open implies p(A) open.

p is a quotient map if and only if p is surjective,

continuous, and A saturated and closed implies p(A)closed.

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Quotient topology

Quotient topologyLet X be a space, A a set, and p: X! A a surjective map.Then there is exactly one topology on A that makes p a

quotient map, such topology is called quotient topology.

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Quotient topology

Quotient topologyLet X be a space, A a set, and p: X! A a surjective map.Then there is exactly one topology on A that makes p a

quotient map, such topology is called quotient topology.

Remark

One must check that:

fi = fU A j p 1(U) is open in Xg

is a topology on A.

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Partitions

Denition (Partition)

Let X be a set. A partition X of X is a collection ofdisjoint subsets of X with union X

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Partitions

Denition (Partition)

Let X be a set. A partition X of X is a collection ofdisjoint subsets of X with union X

Quotient space

Let X be a topological space, X be a partition of X, andp: X! X the natural surjection. If we give X the quotient

topology induced by p, the space X is called a quotientspace of X

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Quotients and subspaces

If p: X! Y is a quotient map and A X is a subspace, itdoes not necessarily follow that the restriction of p:

A ! p(A) is a quotient map. However we have:

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Quotients and subspaces

If p: X! Y is a quotient map and A X is a subspace, itdoes not necessarily follow that the restriction of p:

A ! p(A) is a quotient map. However we have:

Theorem (Quotients and subspaces)

Let p: X! Y be a quotient map and A X a subspace,that is saturated with respect to p, and let q: A ! p(A)the restriction of p. Then:

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Quotients and subspaces

If p: X! Y is a quotient map and A X is a subspace, itdoes not necessarily follow that the restriction of p:

A ! p(A) is a quotient map. However we have:

Theorem (Quotients and subspaces)

Let p: X! Y be a quotient map and A X a subspace,that is saturated with respect to p, and let q: A ! p(A)the restriction of p. Then:

if A is open or closed, then q is a quotient map,

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Quotients and subspaces

If p: X! Y is a quotient map and A X is a subspace, itdoes not necessarily follow that the restriction of p:

A ! p(A) is a quotient map. However we have:

Theorem (Quotients and subspaces)

Let p: X! Y be a quotient map and A X a subspace,that is saturated with respect to p, and let q: A ! p(A)the restriction of p. Then:

if A is open or closed, then q is a quotient map,

if p is open or closed, then q is a quotient map.

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Quotients and compositions

Theorem

Composition of quotient maps is a quotient map.

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Fundamental theorem

Theorem (Fundamental theorem)

Let p: X! Y be a quotient map. Let Z be a space, andg: X! Z a continuous map that is constant on each berof p. Then g induces a continuous map g: Y! Z such

that g p= g.

X Z

Y

g

p

g

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Fundamental theorem

Theorem (Fundamental theorem, for equivalencerelations)

Let X be a space, and an equivalence relation on X.Let g: X! Z be a continuous map such that g(x) = g(x0)whenever x x0. Then, if we denote by p: X! X= thequotient map, the map g induces a continuous map

g: X=! Z such that g p= g

X Z

X=

g

pg