286: Induced Map from Domain Quotient of Continuous Map Is Continuous|T.B.P.

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A description/proof of that induced map from domain quotient of continuous map is continuous

Topics

About: topology space
About: map

Starting Context

The reader knows a definition of continuous map.
The reader knows a definition of quotient of set.
The reader knows a definition of quotient map.
The reader knows a definition of quotient topology.

Target Context

The reader will have a description and a proof of the proposition that for any continuous map, its induced map (if exists) from any domain quotient is continuous.

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

Main Body

1: Description

For any topological spaces,

M_{1}

and

M_{2}

, any continuous map,

f : M_{1} \to M_{2}

, and any quotient of

M_{1}

M_{1} / \sim

, and the quotient map,

ρ : M_{1} \to M_{1} / \sim

, if f equals for all the elements of

M_{1}

that correspond to each element of

M_{1} / \sim

, there is the induced map,

\tilde{f}

, as in

f : M_{1} \to_{ρ}^{} M_{1} / \sim \to_{\tilde{f}}^{} M_{2}

, and

\tilde{f}

is continuous.

2: Proof

For any open set,

U \subseteq M_{2}

f^{- 1} (U)

is open.

f^{- 1} (U) = ρ^{- 1} \circ {\tilde{f}}^{- 1} (U)

, because for any

p \in f^{- 1} (U)

\tilde{f} \circ ρ (p) \in U

ρ (p) \in {\tilde{f}}^{- 1} (U)

p \in ρ^{- 1} \circ {\tilde{f}}^{- 1} (U)

; on the other hand, for any

p \in ρ^{- 1} \circ {\tilde{f}}^{- 1} (U)

ρ (p) \in {\tilde{f}}^{- 1} (U)

\tilde{f} \circ ρ (p) \in U

where

\tilde{f} \circ ρ (p) = f (p)

, so,

p \in f^{- 1} (U)

. By the definition of quotient topology, if

ρ^{- 1} \circ {\tilde{f}}^{- 1} (U)

is open (in fact, it is),

{\tilde{f}}^{- 1} (U)

is open, so, as for any open set,

U \subseteq M_{2}

{\tilde{f}}^{- 1} (U)

is open,

{\tilde{f}}^{- 1}