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Let $G$ be a connected algebraic group. Then the union of all of its Borel subgroups is $G$ itself. I am following this proof on page 70 lemma 26.3.

Let's just focus on the first part proving the union of all Borel subgroups is closed. But I found the proof is a kinda flawed. Let $B$ be a Borel subgroup. Consider $$G\times B\xrightarrow{\phi} G\times G\xrightarrow{\pi} G/B\times G\xrightarrow{\gamma} G$$ where $\phi:(g,b)\mapsto(g,gbg^{-1})$, $\pi:(g,h)\mapsto(gB,h)$, $\gamma:(gB,h)\mapsto h$.

In total, the image is the union of the conjugates of the Borel subgroup. We would like to show it is closed. Then we want to show $im(\pi\phi)$ is closed then by $G/B$ projectivity, we could conclude. But then I ran into trouble of being convinced by the proof.

First, he wants to show $im(\phi)$ is closed. He cited a result proved before, corollary 16.5, which claims that the image of an algebraic group homomorphism is an algebraic group (closed). But this $\phi$ is not a group homomorphism, because $\phi((g,h)(x,y))=\phi(gx,hy)=(gx,gxhyx^{-1}g^{-1})$ but $\phi((g,h))\phi((x,y))=(g,ghg^{-1})(x,xyx^{-1})=(gx,ghg^{-1}xyx^{-1})$. Apparently, they are not equal...

Second, even if assume we have got $im(\phi)$ is closed, we want the image under $\pi$ is closed but $\pi$ is an open (surjective) map which was also proved. However, I do not think open surjective map implies closed map.

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Consider the map $G\times G\to G\times G$ given by $(g,g')\mapsto (g,gg'g^{-1})$. This is an isomorphism with inverse given by $(x,y)\mapsto (x,x^{-1}yx)$. So, since $G\times B$ is a closed subset of $G\times G$, the image in $G\times G$ is closed.

Why the image of $\mathrm{im}(\phi)$ under $\pi$ is closed is slightly more sophisticated. Namely, note that $(G/B)\times G$ is just $(G\times G)/(B\times 1)$. Now, since $G\times G\to (G/B)\times G$ is a quotient map, to check a subset of $(G/B)\times G$ is closed it suffices to check the preimage is closed in $G\times G$. But, $\mathrm{im}(\phi)$ is quickly checked to be $B\times 1$ stable, so $\pi^{-1}(\pi(\mathrm{im}(\phi)))=\mathrm{im}(\phi)$.

NB: I'm using classical language since this is what Szamuely is using. Let me know if you need it translated into schemes.

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  • $\begingroup$ Thank you! Last question why it is enough to check the preimage is closed for $G\times G\to G/B\times G$? The map is a quotient map: open surjective continuous. But somehow I have in my mind that an open surjective continuous map need not be closed? Say $\pi:X\to Y$ is open surjective continuous. $U$ is closed, if $\pi^{-1}(U)$ is closed implies $U$ is closed for any $U$. Does it implies that it is a closed map? $\endgroup$
    – CO2
    Aug 15, 2020 at 22:55
  • $\begingroup$ Because a subset of a quotient space is closed iff its preimage is closed. That's not the same as being a closed map. $\endgroup$ Aug 15, 2020 at 22:58
  • $\begingroup$ Ah you are right. We learnt $U$ is open if and only if $\pi^{-1}U$ is open for a quotient map. But the pullback commutes with complement So also true for closed sets. It is always a bit confusing when it relates to open and closed maps... Also I think it would be nice to see the scheme way. I am learning them right now. Thank you. $\endgroup$
    – CO2
    Aug 15, 2020 at 23:04

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