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The $f$-localization I mean is the one described and studied in detail in the book by E. D. Farjoun; $L_f$ is a homotopy idempotent functor which associates to each space $X$ an $f$-equivalence $X\to L_f(X)$ where $L_f(X)$ is $f$-local.

$f$-localization has a kind of uniqueness: if $F$ is some other coaugmented functor with the property that $F(X)$ is $f$-local for every $X$ (I'm happy to assume that $F = L_g$ for some map $g$), then there is a commutative square of functors and natural transformations, which I don't know how to draw here. The square would show that the composites $$ id \xrightarrow{\iota} L_f \xrightarrow{L_f(j)} L_f\circ F \qquad \mathrm{and} \qquad id \xrightarrow{j} F \xrightarrow{\iota_F} L_f \circ F $$ are equal. And $\iota_F$ is a weak equivalence for every space $X$; thus $F$ factors through $L_f$ `up to weak equivalence'.

My Question: Suppose $X\to Y$ is an $f$-equivalence; then $L_f(X) \to L_f(Y)$ is a weak equivalence; does it follow that $(L_f\circ F)(X) \to (L_f\circ F)(Y)$ is a weak equivalence?

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If you allow $F$ to be a continuous functor, which it appears you do since localization functors are continuous, then I believe this is true. Corollary 2.1.1 in the following paper seems to be useful:

E.Dror Farjoun, Higher homotopies of natural constructions, Journal of Pure and Applied Algebra, Volume 108, Issue 1, 8 April 1996, Pages 23-34, ISSN 0022-4049,

It says that if $\phi: X \rightarrow Y$ is a map and $G$ is a continuous, coaugmented, idempotent functor such that for all spaces $A$, the induced map $map(Y,GA) \rightarrow map(X,GA)$ is a weak equivalence, then $G(\phi): G(X) \rightarrow G(Y)$ is a weak equivalence. Letting $G = L_f \circ F$ should give the result you want. The induced map will be an equivalence since $\phi$ is an $f$-equivalence and $L_f\circ F(A)$ is $f$-local for all spaces $A$.

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I think that in general the answer is `no', since your functor $F$ need not preserve $f$-equivalences. Take $f\colon \ast \to \ast$ to be the identity map or any other weak equivalence, then the $f$-local objects are the fibrant objects and $L_f$ is equivalent to the fibrant replacement is simplicial sets. Now take any functor $F$ which does not preserve weak equivalences and takes fibrant values. For example, $F(-) = \mathrm{fib}(\mathrm{hom}(A,-))$, where $\mathrm{fib}$ is a fibrant replacement functor and $A$ a non-trivial space. Then $L_f\circ F$ does not preserve weak equivalences (which are also $f$-equivalences) of non-fibrant simplicial sets.

I think that the condition that you need to assume about $F$ is that $F$ preserves $f$-equivalences.

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    $\begingroup$ I really like this example. I never considered this $F$ before, but it does exactly what you want. I also agree that in situations like this the condition you suggest sounds right $\endgroup$
    – David White
    Commented May 1, 2013 at 15:46
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An $f$-equivalence is a map $\alpha:X\to Y$ which induces a weak equivalence $\alpha^\ast:map_\ast(Y,Q)\to map_\ast (X,Q)$ for all $f$-local spaces $Q$; equivalently, $L_f(\alpha)$ is a weak equivalence. Therefore the assumption implies that every $g$-local space is $f$-local, so an $f$-equivalence $\alpha$ is also a $g$-equivalence; so $L_g(\alpha)$ is a weak equivalence.

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    $\begingroup$ So this answers it for $F=L_g$, and Matthew Sartwell's answer is slightly more general, but Boris shows you can't go more general than that. Interesting. +1 to everyone $\endgroup$
    – David White
    Commented May 1, 2013 at 15:36

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