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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}

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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.

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

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