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One can prove the following:

Let $X,Y$ are topological spaces, $f: X \to Y$ continuous and surjective and $Y$ with the identification topology induced by $f$. If $B \subseteq Y$ is such that $f^{-1}(B) = A$ is closed then $B$ with the subspace topology has the identification topology induced by $f|_A: A \to B$.

I believe I proved the following also:

Let $X,Y$ are topological spaces, $f: X \to Y$ continuous and surjective and $Y$ with the identification topology induced by $f$. If $B \subseteq Y$ is such that $f^{-1}(B) = A$ is open then $B$ with the subspace topology has the identification topology induced by $f|_A: A \to B$.

Could somebody confirm that this result really holds? Basically, the proof of the first lemma in terms of closed sets can be used by replacing all the occurrences of closed by open.

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Yes, the result is correct. I don’t know exactly how you proved the first result, so I can’t be absolutely sure that the replacement that it translates as you suggest, but I think it likely that it does. I’d argue as follows. First, $B$ is open in $Y$ by the definition of the quotient topology. Now let $U\subseteq B$. Then $U$ is open in $B$ iff $U$ is open in $Y$ iff $f^{-1}[U]$ is open in $X$ iff $(f\upharpoonright A)^{-1}[U]$ is open in $A$, since $f^{-1}[U]\subseteq A$ and $A$ is open in $X$.

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Yes, this says that the restriction $f|_A:A\to B$ of the quotient map $f:X\to Y$ is again a quotient map if $A$ is saturated and open (or saturated and closed). More generally, we have the following result:

Let $f:X\to Y$ be a quotient map and $A\subset X$ with $B=f[A]$, where $A$ and $B$ have the subspace topologies. Then $f|_A:A\to B$ is a quotient map if and only if:
(i) Each subset of $A$ which is saturated and open in $A$ is the intersection of $A$ with a saturated and open subset of $X$.
(ii) The same with 'open' replaced by 'closed'.

Then your result follows immediately since a subset of $A$ which is saturated and open in $A$ is also saturated and open in $X$ if $A$ itself is saturated and open.

One application of the above lemma is when $X$ is a space, $A$ is a saturated subspace, and $f$ is injective on $X\setminus A$. We can then show that a saturated closed subset $C$ of $A$ has a saturated closure such that $\text{cl}(C)\cap A=C$. So in that case $f|_A$ is a quotient map.

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