# 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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## 1 Answer

No. Consider the case where your sets happen to be abelian groups with an action of G. Pushing out the zero morphism over any map gives the cokernel, so a special case of your question is whether taking G-invariants is right exact, and it isn't. This should give an idea of where counterexamples might come from.

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.

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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
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