Open and closed maps

Due to the parallel definitions and other similarities there is little point to having separate pages.

Definition

Open map

A

$$f:(X,\mathcal{J})\rightarrow (Y,\mathcal{K})$$ (which need not be continuous) is said to be an open map if:

• The image of an open set is open (that is $$\forall U\in\mathcal{J}[f(U)\in\mathcal{K}]$$ )

Closed map

A

$$f:(X,\mathcal{J})\rightarrow (Y,\mathcal{K})$$ (which need not be continuous) is said to be a closed map if:

• The image of a closed set is closed

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TODO: References - it'd look better

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Importance with respect to the quotient topology

The primary use of recognising an open/closed map comes from recognising a Quotient topology, as the following theorem shows

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Theorem: Given a map

$$f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})$$ that is continuous, surjective and an open or closed map, then the topology $\mathcal{K}$ on $Y$ is the same as the Quotient topology induced on $Y$ by $f$

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Note: the intuition behind this theorem is fairly obvious, if $f$ is an open map then it takes open sets to open sets, but the quotient topology is precisely those sets in $Y$ whose preimage is open in $X$

Using the first theorem, automatically we see that $K$ is weaker than the quotient topology (that is $\mathcal{K}\subseteq\mathcal{Q}$)

To show equality we need only to show $\mathcal{Q}\subseteq\mathcal{K}$

Open case:

Suppose that $f$ is a continuous open map

Let $A\in\mathcal{Q}$ be given

we need to show this $\implies A\in\mathcal{K}$ then we use the Implies and subset relation to get what we need.

This means $f^{-1}(A)\in\mathcal{J}$ but as $f$ is an open mapping $\forall U\in\mathcal{J}[f(U)\in\mathcal{K}]$ so:

$f(f^{-1}(A))\in\mathcal{K}$ (it is open in $Y$)

Warning: we have not shown that $A\in\mathcal{K}$ yet!

Because $f$ is surjective we know $f(f^{-1}(A))=A$ (as for all things in $A$ there must be something that maps to them, by surjectivity)

So we conclude $A\in \mathcal{K}$

So $A\in\mathcal{Q}\implies A\in\mathcal{K}$ or $\mathcal{Q}\subseteq\mathcal{K}$

Combining this with $\mathcal{K}\subseteq\mathcal{Q}$ we see:

$\mathcal{K}=\mathcal{Q}$

Closed case:

Suppose $f$ is a continuous closed map

Let $A\in\mathcal{Q}$ be given

we need to show this $\implies A\in\mathcal{K}$ then we use the Implies and subset relation to get what we need.

Obviously, because $f$ is continuous $f^{-1}(A)\in\mathcal{J}$ - that is to say $f^{-1}(A)$ is open in $X$

That means that $X-f^{-1}(A)$ is closed in $X$ - just by definition of an open set

Since $f$ is a closed map, $f(X-f^{-1}(A))$ is also closed

BUT! $f(X-f^{-1}(A))=Y-A$ by surjectivity (as the set $X$ take away ALL the things that map to points in $A$ will give you all of $Y$ less the points in $A$)

As $f(X-f^{-1}(A))$ is closed and equal to $Y-A$, $Y-A$ is closed

this means $Y-(Y-A)$ is open, but $Y-(Y-A)=A$ so $A$ is open.

That means $A\in\mathcal{K}$

We have shown $A\in\mathcal{Q}\implies A\in\mathcal{K}$ and again by the Implies and subset relation we see $\mathcal{Q}\subseteq\mathcal{K}$

Combining this with $\mathcal{K}\subseteq\mathcal{Q}$ we see:

$\mathcal{K}=\mathcal{Q}$

This completes the proof.

See also

• Continuous map
• Topology