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Feb 15, 2015 at 2:40 comment added Eric Wofsey Another way to say this without using any model category machinery is that $|K(f)|$ and $|K(g)|$ become homotopic after composing with the map $|K(Y)|\to Y$. But this latter map is a weak equivalence and hence induces a bijection on homotopy classes of maps out of any CW-complex, hence $|K(f)|$ and $|K(g)|$ must be homotopic.
Feb 15, 2015 at 1:05 comment added pw1822 Could you explain what do you do you mean by ,,$|K(f)|$ and $|K(g)|$ are wekaly homotopic? It seems to me as if it was connected with model categrories, am I right? But I do not understand very well this stuff yet.
Feb 15, 2015 at 0:57 vote accept pw1822
Feb 15, 2015 at 0:55 history answered Eric Wofsey CC BY-SA 3.0