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If $\mathcal M$ is a model category and I know that $A$ and $B$ are isomorphic in $\mathrm{Ho}(\mathcal M)$, is it guaranteed that there is a zig-zag of weak-equivalences in $\mathcal M$ connecting $A$ and $B$?

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  • $\begingroup$ In particular, you could have searched google.com/… . If you click the top link, it has the answer. $\endgroup$
    – Harry Gindi
    Commented Mar 23, 2010 at 5:06

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Yes. The isomorphism in $\mathrm{Ho}(\mathcal{M})$ is represented by a morphism in $\mathcal{M}$ from a cofibrant replacement for $A$ to a fibrant replacement for $B$. The "converse to the Whitehead lemma" states that a map in a model category is a weak equivalence iff its image in the homotopy category is an isomorphism. Combining this with the definition of (co)fibrant replacement, we see that $A$ and $B$ are connected by a 3-step zig-zag of weak equivalences.

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