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A profinite group is said to be projective if its cohomological dimension is $\leq 1$. Is this related to some other notion of "projective"? How so?

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A profinite group $P$ is projective if and only if any continuous group homomorphism from it to a profinite quotient group $G/H$ lifts to a continuous group homomorphism to the profinite group $G$.

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I think you mean any continuous group epimorphism from it to a profinite quotient group $G/H$ lifts to a continuous group homomorphism to the profinite group $G$. Are you trying to talk about a similarity with projective modules? If so, we should explain why it is an epimorphism we're talking about and not a homomorphism. –  Makhalan Duff Nov 2 '10 at 13:16
    
You should take another look at the definition of projective module. Only the map from G to G/H needs to be an epi, not the map from P to G/H. If I interpret Leonid Positselski correctly, he is simply saying that P is a projective object in the category of profinite groups. –  Saul Glasman Nov 2 '10 at 13:45
    
Ah, yes. You're correct. Very good! –  Makhalan Duff Nov 2 '10 at 14:21
    
@Saul: I am not quite sure about what should be meant by a projective object in a category, except if the category is abelian. One can imitate the conventional definition for abelian categories, but then one has to explain what is meant by an epimorphism in a category. Once again, there is the standard definition, but it is not applicable in the case at hand. Not every categorical epimorphism in the category of (say) finite noncommutative groups is a surjective map. If one replaces (categorical) epimorphisms with universal epimorphisms, it will finally work. –  Leonid Positselski Nov 2 '10 at 20:10
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It is explained in Section 7.6 of Ribes and Zalesskii, "Profinite groups". The notion is similar to the notion of a projective module. For example, free profinite groups are projective. Moreover, a profinite group is projective if and only if it is a closed subgroup of a free profinite group.

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