# Does the covering involution lift to the pullback cover?

Let $$X$$ and $$Y$$ be smooth projective varieties over $$\mathbb{C}$$ such that $$p:X \rightarrow Y$$ is an etale Galois double cover.

For a smooth projective variety $$Z$$ with $$f: Z \rightarrow Y$$ the pullback gives an induced etale double cover $$P: W \rightarrow Z$$ with $$W:=Z\times_Y X$$ and the map $$F: W\rightarrow X$$.

Now there is the covering involution $$\iota: X \rightarrow X$$ with $$\iota^2=id_X$$ and $$p\circ \iota=p$$.

$$\textbf{Question:}$$ Does the covering involution lift to the double cover $$P:W \rightarrow Z$$?

We get a morphism $$I: W\times_X X\rightarrow W$$ using $$\iota: X\rightarrow X$$ and $$F: W\rightarrow X$$.

Identifying $$W\times_X X$$ and $$W$$ using the canonical isomorphism $$W\cong W\times_X X$$, is the morphism $$I$$ the (a) covering involution of $$P: W \rightarrow Z$$, i.e. do we have $$I^2=id_W$$ and $$P\circ I=P$$? And does $$I$$ live over $$\iota$$, i.e. do we have the following cartesian diagramm(s): $$\require{AMScd}$$ $$\begin{CD} W @>I>> W@>P>> Z\\ @V F V V @VV F V@VV f V\\ X @>>\iota> X @>>p> Y \end{CD}$$

Yes to all the questions. They all follow from the universal property of the fiber product. Except that your definition of $$I$$ needs to be explained a bit.

But first, let us give another construction of $$I$$. We will prove first that this other construction answers all your question (easily), then we will discuss a bit your construction. For this, consider the map $$\iota\circ F:W\to X\to X$$ and the map $$P:W\to Z$$. If we compose with $$p$$ and $$f$$ we get the same maps : $$p\circ \iota\circ F=p\circ F=f\circ P$$. Thus by universal property of fiber product, the pair of maps $$(\iota\circ F, P)$$ gives a unique map $$I:W\to W$$ such that $$P\circ I=P$$ and $$F\circ I=\iota\circ F$$.

We thus have a commutative diagram $$\require{AMScd} \begin{CD} W@>I>>W@>I>>W@>P>>Z\\ @VFVV@VFVV@VFVV@VVV\\ X@>\iota>>X@>\iota>>X@>>>Y \end{CD}$$ Since $$\iota\circ\iota=\operatorname{id}_X$$, then the map $$I^2:W\to W$$ satisfy $$F\circ I^2=F$$ and $$P\circ I^2=P$$. Since the identity of $$W$$ also satisfy this requirements, by unicity in the universal property $$I^2=\operatorname{id}_W$$.

This answer your question with this map $$I$$.

Now, let us discuss your construction of $$I$$. Note that you should be careful with the canonical isomorphism $$W\times_X X\simeq W$$ because in this fiber product, $$X\to X$$ is not the identity. In fact, if $$X\to X$$ was not an isomorphism, $$W\times_X X$$ would not be isomorphic to $$W$$... We should actually write the map here.

So what is "this natural isomorphism" $$W\times_{X,\iota} X\simeq W$$. Well there is the projection onto the first factor. This is indeed an isomorphism, but NOT the one you want. There is another one, induced by the chain of isomorphisms $$W\times_{X,\iota} X=(Z\times_Y X)\times_{X,\iota} X\simeq Z\times_Y (X\times_{X,\iota} X)\simeq Z\times_{Y,\iota}X\simeq Z\times_Y X=W$$. I let you write concretely what is this isomorphism. This is the one you want...

Now, to prove that this is the same as the $$I$$ defined in the first part, it is enough to check that it same universal property.

• Thanks, this is a very good answer and helps a lot. And thank you for the hint about the "canonical" isomorphism! – Bernie Feb 14 at 9:01