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Let $f:X \rightarrow Y$ be a mapping of a topological space $X$ into a set $Y$. Let $J$ be the topology on $X$. What exactly is "the topology carried over to $Y$ by $f$"?

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up vote 2 down vote accepted

There is a final topology determined by $f$: here the open sets of $Y$ are the subsets $V \subset Y$ such that $f^{-1}(V)$ is open in $X$. This satisfies a universal property: see Proposition 1.7.1 of Schapira's notes. Assuming you want $f$ to be continuous, this topology is as fine as it could be in the sense that it contains any topology which makes $f$ continuous.

If $f$ is surjective then we've just described the quotient topology determined by $f$. I must admit that I can't think of any good examples in which this is not the case.

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Would you be willing to give more context, by the way? We might be able to figure this out for sure, then. – Dylan Moreland Apr 27 '12 at 18:42
Hi Dylan, i think its exactly what you explained. Can you explain a little bit what do you mean "as fine as it can be"? – Manos Apr 27 '12 at 18:46
A topology gets "finer" the more open sets it has. If there were any open sets in $Y$ besides those already described, $f$ wouldn't be continuous. – Thomas Belulovich Apr 27 '12 at 18:48
Exactly, I claim that any topology on $Y$ for which $f$ is continuous is contained in this one. – Dylan Moreland Apr 27 '12 at 18:58
Actually, I found a more common term for it. Will edit. – Dylan Moreland Apr 27 '12 at 18:59

That topology consists of exactly the subsets $U\subseteq Y$ for which $f^{-1}(U) \in J$, that is $f^{-1}(U)$ is open in $X$.

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One should also emphasise the universal property: with the final topology on $Y$, if $Z$ is a topological space, then a function $g:Y \to Z$ is continuous if and only if the composition $gf$ is continuous. Many constructions in general topology are designed to enable the construction of continuous functions. The final topology is also called the identification topology in the case $f$ is surjective.

Dually, there is an initial topology for $f:X \to Y$ if now $X$ is a set, and $Y$ is a topological space.

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