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Tommaso Rossi
Tommaso Rossi
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In what follows all the groups will be discrete, not necessarly finite.

Let $f:G\to H$ be a morphism of groups and $H'\to H$ be the inclusion of a subgroup. It seems to me (but correct me if I am wrong) that the pullback $BG\times_{BH}BH'$ is $B(f^{-1}H)$$B(f^{-1}H')$.

If this is true, is there a more general statement that tells you when a commutative square of classifying space is a pullback diagram?

In what follows all the groups will be discrete, not necessarly finite.

Let $f:G\to H$ be a morphism of groups and $H'\to H$ be the inclusion of a subgroup. It seems to me (but correct me if I am wrong) that the pullback $BG\times_{BH}BH'$ is $B(f^{-1}H)$.

If this is true, is there a more general statement that tells you when a commutative square of classifying space is a pullback diagram?

In what follows all the groups will be discrete, not necessarly finite.

Let $f:G\to H$ be a morphism of groups and $H'\to H$ be the inclusion of a subgroup. It seems to me (but correct me if I am wrong) that the pullback $BG\times_{BH}BH'$ is $B(f^{-1}H')$.

If this is true, is there a more general statement that tells you when a commutative square of classifying space is a pullback diagram?

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Tommaso Rossi
Tommaso Rossi
  • 641
  • 3
  • 10

Pullbacks of classifying spaces

In what follows all the groups will be discrete, not necessarly finite.

Let $f:G\to H$ be a morphism of groups and $H'\to H$ be the inclusion of a subgroup. It seems to me (but correct me if I am wrong) that the pullback $BG\times_{BH}BH'$ is $B(f^{-1}H)$.

If this is true, is there a more general statement that tells you when a commutative square of classifying space is a pullback diagram?