# Fixed points sets of pushouts

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?

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Concretely, lets say your group is $\mathbb{Z}/2$ and you push out $\mathbb{Z}/2 \to \ast$ over itself. The ordinary pushout is a point (with the trivial action). But if you take invariants first you find you are pushing two points out over the empty set, which results in two points.
It might also be worth pointing out that taking $G$ invariants is where group cohomology comes from: it measures the failure of $(-)^G$ to be right-exact. – Bill Kronholm Apr 21 '10 at 12:29