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Let $G$ be a group and $X \to Y, X \to Z$ morphisms of $G$-sets with pushout $P=Y \cup_X Z$. Is then $P^G$ the pushout of $X^G \to Y^G, X^G \to Z^G$? This is not clear from general category theory, because pushouts do not have to commute with equalizers. I hope it's true when $X \to Y$ is injective (thus also $Z \to P$). Then $Y^G \cup_{X^G} Z^G \to P^G$ is injective, but I don't know why an element of $P^G$ which comes from $Y$ lies in the image...

Now let's work with $G$-spaces instead of $G$-sets. Then it is not clear at all to me if the canonical map from the pushout to $P^G$ is open. Is it true when $G$ is finite? Or if $X,Y,Z$ are $G$-CW-complexes? By the way, this was used in a lecture about how to compute the homology of the classifying space $BG$ with the help of $\underline{E} G$.

Let $G$ be a group and $X \to Y, X \to Z$ morphisms of $G$-sets with pushout $P=Y \cup_X Z$. Is then $P^G$ the pushout of $X^G \to Y^G, X^G \to Z^G$? This is not clear from general category theory, because pushouts do not have to commute with equalizers. I hope it's true when $X \to Y$ is injective (thus also $Z \to P$). Then $Y^G \cup_{X^G} Z^G \to P^G$ is injective, but I don't know why an element of $P^G$ which comes from $Y$ lies in the image...

Now let's work with $G$-spaces instead of $G$-sets. Then it is not clear at all to me if the canonical map from the pushout to $P^G$ is open. Is it true when $G$ is finite? Or if $X,Y,Z$ are $G$-CW-complexes? By the way, this was used in a lecture about how to compute the homology of the classifying space $BG$ with the help of $\underline{E} G$.

EDIT: I was told that it is true if $X,Y;Z$ are $G$-CW-complexes and $X \to Y$ is a $G$-cofibration. How is this proven?

Let $G$ be a group and $X \to Y, X \to Z$ morphisms of $G$-sets with pushout $P$. Is then $P^G$ the pushout of $X^G \to Y^G, X^G \to Z^G$? This is not clear from general category theory, because pushouts do not have to commute with equalizers. I hope it's true when $X \to Y$ is injective (thus also $Z \to P$). Then $Y^G \cup_{X^G} Z^G \to P^G$ is injective, but I don't know why an element of $P^G$ which comes from $Y$ lies in the image...

Now let's work with $G$-spaces instead of $G$-sets. Then it is not clear at all to me if the canonical map from the pushout to $P^G$ is open. Is it true when $G$ is finite? Or if $X,Y,Z$ are $G$-CW-complexes? By the way, this was used in a lecture about how to compute the homology of the classifying space $BG$ with the help of $\underline{E} G$.

Let $G$ be a group and $X \to Y, X \to Z$ morphisms of $G$-sets with pushout $P=Y \cup_X Z$. Is then $P^G$ the pushout of $X^G \to Y^G, X^G \to Z^G$? This is not clear from general category theory, because pushouts do not have to commute with equalizers. I hope it's true when $X \to Y$ is injective (thus also $Z \to P$). Then $Y^G \cup_{X^G} Z^G \to P^G$ is injective, but I don't know why an element of $P^G$ which comes from $Y$ lies in the image...

Now let's work with $G$-spaces instead of $G$-sets. Then it is not clear at all to me if the canonical map from the pushout to $P^G$ is open. Is it true when $G$ is finite? Or if $X,Y,Z$ are $G$-CW-complexes? By the way, this was used in a lecture about how to compute the homology of the classifying space $BG$ with the help of $\underline{E} G$.

Let $G$ be a group and $X \to Y, X \to Z$ morphisms of $G$-sets with pushout $P$. Is then $P^G$ the pushout of $X^G \to Y^G, X^G \to Z^G$? This is not clear from general category theory, because pushouts do not have to commute with equalizers. I think it's true when $X \to Y$ is injective; my proof is a fiddly computation, perhaps someone can provide a nice proof?

Now let's work with $G$-spaces instead of $G$-sets. Then it is not clear at all to me if the canonical map from the pushout to $P^G$ is open. Is it true when $G$ is finite? Or if $X,Y,Z$ are $G$-CW-complexes? By the way, this was used in a lecture about how to compute the homology of the classifying space $BG$ with the help of $\underline{E} G$.

Let $G$ be a group and $X \to Y, X \to Z$ morphisms of $G$-sets with pushout $P$. Is then $P^G$ the pushout of $X^G \to Y^G, X^G \to Z^G$? This is not clear from general category theory, because pushouts do not have to commute with equalizers. I hope it's true when $X \to Y$ is injective (thus also $Z \to P$). Then $Y^G \cup_{X^G} Z^G \to P^G$ is injective, but I don't know why an element of $P^G$ which comes from $Y$ lies in the image...

Now let's work with $G$-spaces instead of $G$-sets. Then it is not clear at all to me if the canonical map from the pushout to $P^G$ is open. Is it true when $G$ is finite? Or if $X,Y,Z$ are $G$-CW-complexes? By the way, this was used in a lecture about how to compute the homology of the classifying space $BG$ with the help of $\underline{E} G$.