$forall, 0to Nto Mto Fto 0 $ exact, $forall R$-module $E$, $F$ is flat $implies 0 to Notimes Eto Motimes Eto...












2












$begingroup$


In $R$-mod category, $forall, 0to Nto Mto Fto 0 $ is short exact sequence, $forall R$-module $E$.



$forall, 0to Nto Mto Fto 0 $ exact, $forall R$-module $E$, $F$ is flat $implies 0 to Notimes Eto Motimes Eto Fotimes E to0$ is exact.



How to prove $F$ is flat $implies$$quad0 to Notimes Eto Motimes Eto Fotimes E to0$ is exact?



My thought:



I gave an answer. I wonder if there's an alternative and more direct proof, just using definition of flat module instead of introducing another short exact sequence, since it's not easy for beginners to think of that method. Thanks in advance :)






abstract-algebra modules homological-algebra exact-sequence flatness







share|cite|improve this question











$endgroup$












  • $begingroup$
    $K=E{}{}{}{}{}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 9:07










  • $begingroup$
    @LordSharktheUnknown Yes, I put it right.
    $endgroup$
    – Andrews
    Dec 24 '18 at 9:14










  • $begingroup$
    I don't think you mean to say that $(cdot otimes E)$ is an exact functor in the original title because it seems it only preserves exactness of a specific kind of short exact sequence.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 11:00










  • $begingroup$
    @AniruddhAgarwal Yes, you're right. I intended to shorten the title but made a mistake. Thanks for mention that.
    $endgroup$
    – Andrews
    Dec 24 '18 at 11:32
















2












$begingroup$


In $R$-mod category, $forall, 0to Nto Mto Fto 0 $ is short exact sequence, $forall R$-module $E$.



$forall, 0to Nto Mto Fto 0 $ exact, $forall R$-module $E$, $F$ is flat $implies 0 to Notimes Eto Motimes Eto Fotimes E to0$ is exact.



How to prove $F$ is flat $implies$$quad0 to Notimes Eto Motimes Eto Fotimes E to0$ is exact?



My thought:



I gave an answer. I wonder if there's an alternative and more direct proof, just using definition of flat module instead of introducing another short exact sequence, since it's not easy for beginners to think of that method. Thanks in advance :)






abstract-algebra modules homological-algebra exact-sequence flatness







share|cite|improve this question











$endgroup$












  • $begingroup$
    $K=E{}{}{}{}{}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 9:07










  • $begingroup$
    @LordSharktheUnknown Yes, I put it right.
    $endgroup$
    – Andrews
    Dec 24 '18 at 9:14










  • $begingroup$
    I don't think you mean to say that $(cdot otimes E)$ is an exact functor in the original title because it seems it only preserves exactness of a specific kind of short exact sequence.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 11:00










  • $begingroup$
    @AniruddhAgarwal Yes, you're right. I intended to shorten the title but made a mistake. Thanks for mention that.
    $endgroup$
    – Andrews
    Dec 24 '18 at 11:32














2












2








2





$begingroup$


In $R$-mod category, $forall, 0to Nto Mto Fto 0 $ is short exact sequence, $forall R$-module $E$.



$forall, 0to Nto Mto Fto 0 $ exact, $forall R$-module $E$, $F$ is flat $implies 0 to Notimes Eto Motimes Eto Fotimes E to0$ is exact.



How to prove $F$ is flat $implies$$quad0 to Notimes Eto Motimes Eto Fotimes E to0$ is exact?



My thought:



I gave an answer. I wonder if there's an alternative and more direct proof, just using definition of flat module instead of introducing another short exact sequence, since it's not easy for beginners to think of that method. Thanks in advance :)






abstract-algebra modules homological-algebra exact-sequence flatness







share|cite|improve this question











$endgroup$




In $R$-mod category, $forall, 0to Nto Mto Fto 0 $ is short exact sequence, $forall R$-module $E$.



$forall, 0to Nto Mto Fto 0 $ exact, $forall R$-module $E$, $F$ is flat $implies 0 to Notimes Eto Motimes Eto Fotimes E to0$ is exact.



How to prove $F$ is flat $implies$$quad0 to Notimes Eto Motimes Eto Fotimes E to0$ is exact?



My thought:



I gave an answer. I wonder if there's an alternative and more direct proof, just using definition of flat module instead of introducing another short exact sequence, since it's not easy for beginners to think of that method. Thanks in advance :)






abstract-algebra modules homological-algebra exact-sequence flatness




abstract-algebra modules homological-algebra exact-sequence flatness






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 12:38







Andrews

















asked Dec 24 '18 at 8:54









AndrewsAndrews

1,2812422




1,2812422












  • $begingroup$
    $K=E{}{}{}{}{}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 9:07










  • $begingroup$
    @LordSharktheUnknown Yes, I put it right.
    $endgroup$
    – Andrews
    Dec 24 '18 at 9:14










  • $begingroup$
    I don't think you mean to say that $(cdot otimes E)$ is an exact functor in the original title because it seems it only preserves exactness of a specific kind of short exact sequence.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 11:00










  • $begingroup$
    @AniruddhAgarwal Yes, you're right. I intended to shorten the title but made a mistake. Thanks for mention that.
    $endgroup$
    – Andrews
    Dec 24 '18 at 11:32


















  • $begingroup$
    $K=E{}{}{}{}{}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 9:07










  • $begingroup$
    @LordSharktheUnknown Yes, I put it right.
    $endgroup$
    – Andrews
    Dec 24 '18 at 9:14










  • $begingroup$
    I don't think you mean to say that $(cdot otimes E)$ is an exact functor in the original title because it seems it only preserves exactness of a specific kind of short exact sequence.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 11:00










  • $begingroup$
    @AniruddhAgarwal Yes, you're right. I intended to shorten the title but made a mistake. Thanks for mention that.
    $endgroup$
    – Andrews
    Dec 24 '18 at 11:32
















$begingroup$
$K=E{}{}{}{}{}$?
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 9:07




$begingroup$
$K=E{}{}{}{}{}$?
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 9:07












$begingroup$
@LordSharktheUnknown Yes, I put it right.
$endgroup$
– Andrews
Dec 24 '18 at 9:14




$begingroup$
@LordSharktheUnknown Yes, I put it right.
$endgroup$
– Andrews
Dec 24 '18 at 9:14












$begingroup$
I don't think you mean to say that $(cdot otimes E)$ is an exact functor in the original title because it seems it only preserves exactness of a specific kind of short exact sequence.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 11:00




$begingroup$
I don't think you mean to say that $(cdot otimes E)$ is an exact functor in the original title because it seems it only preserves exactness of a specific kind of short exact sequence.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 11:00












$begingroup$
@AniruddhAgarwal Yes, you're right. I intended to shorten the title but made a mistake. Thanks for mention that.
$endgroup$
– Andrews
Dec 24 '18 at 11:32




$begingroup$
@AniruddhAgarwal Yes, you're right. I intended to shorten the title but made a mistake. Thanks for mention that.
$endgroup$
– Andrews
Dec 24 '18 at 11:32










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let me try and make the question more precise. You are given a short exact sequence $0to Mto Nto Fto 0$, with $F$ flat and you want to prove that, for every $E$, the sequence $0to Motimes Eto Notimes Eto Fotimes Eto 0$ is exact.



Take an exact sequence $0to Xto Yto Eto 0$ with $Y$ projective and build the diagram with exact rows and columns
$$require{AMScd}
begin{CD}
{} @. {} @. {} @. 0 \
@. @. @. @VVV \
{} @. Motimes X @>a>> Notimes X @>b>> Fotimes X @>>> 0 \
@. @VcVV @VdVV @VeVV @. \
0 @>>> Motimes Y @>f>> Notimes Y @>g>> Fotimes Y @>>> 0 \
@. @VhVV @ViVV @VjVV @. \
{} @. Motimes E @>k>> Notimes E @>l>> Fotimes E @>>> 0 \
@. @VVV @VVV @VVV @. \
{} @. 0 @. 0 @. 0
end{CD}
$$
where we have used the fact that $F$ and $Y$ are flat.



Here's the diagram chasing. Let $xinker k$; then $x=h(y)$ and $if(y)=kh(y)=0$, so $f(y)inker i$; hence $f(y)=d(z)$. Since $eb(z)=gd(z)=gf(z)=0$, we conclude $b(z)inker e$, so $b(z)=0$. This implies $z=a(t)$.



Thus $fc(t)=da(t)=d(z)=f(y)$; since $f$ is injective, we have $y=c(t)$ and finally
$$
x=h(y)=hc(t)=0
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for graph chasing. Actually, I want a direct proof without introducing another short exact sequence.
    $endgroup$
    – Andrews
    Dec 24 '18 at 12:39



















0












$begingroup$

We can choose a short exact sequence $0to Kto Lto Eto 0 $ s.t. $L$ is free module, and use properties: $(1),otimes $ is right exact $(2)L$ is free ($implies$flat) $(3) F$ is flat for diagram



$$Notimes Kto Motimes Kto Fotimes K$$
$$downarrow quadquadquadquaddownarrow quadquadquadquaddownarrow$$
$$Notimes Lto Motimes Lto Fotimes L$$
$$downarrow quadquadquadquaddownarrow quadquadquadquaddownarrow$$
$$Notimes Eto Motimes Eto Fotimes E$$
Snake Lemma then yields $quad0 to Notimes Eto Motimes Eto Fotimes E to0$.



Using the same idea, we can show this:



$0 to F' to F to F''$ is exact and $F''$ is flat, then $F'$ is flat $iff F$ is flat.



Especially, for exact sequence $0 to F^0to F^1 to cdots to F^n$, if $F^1, cdots, F^n$ are flat, then $F^0$ is flat.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm confused about how you're applying the snake lemma here.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 10:58










  • $begingroup$
    @AniruddhAgarwal See the other answer and try to find the snake :)
    $endgroup$
    – Andrews
    Dec 25 '18 at 10:44












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Let me try and make the question more precise. You are given a short exact sequence $0to Mto Nto Fto 0$, with $F$ flat and you want to prove that, for every $E$, the sequence $0to Motimes Eto Notimes Eto Fotimes Eto 0$ is exact.



Take an exact sequence $0to Xto Yto Eto 0$ with $Y$ projective and build the diagram with exact rows and columns
$$require{AMScd}
begin{CD}
{} @. {} @. {} @. 0 \
@. @. @. @VVV \
{} @. Motimes X @>a>> Notimes X @>b>> Fotimes X @>>> 0 \
@. @VcVV @VdVV @VeVV @. \
0 @>>> Motimes Y @>f>> Notimes Y @>g>> Fotimes Y @>>> 0 \
@. @VhVV @ViVV @VjVV @. \
{} @. Motimes E @>k>> Notimes E @>l>> Fotimes E @>>> 0 \
@. @VVV @VVV @VVV @. \
{} @. 0 @. 0 @. 0
end{CD}
$$
where we have used the fact that $F$ and $Y$ are flat.



Here's the diagram chasing. Let $xinker k$; then $x=h(y)$ and $if(y)=kh(y)=0$, so $f(y)inker i$; hence $f(y)=d(z)$. Since $eb(z)=gd(z)=gf(z)=0$, we conclude $b(z)inker e$, so $b(z)=0$. This implies $z=a(t)$.



Thus $fc(t)=da(t)=d(z)=f(y)$; since $f$ is injective, we have $y=c(t)$ and finally
$$
x=h(y)=hc(t)=0
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for graph chasing. Actually, I want a direct proof without introducing another short exact sequence.
    $endgroup$
    – Andrews
    Dec 24 '18 at 12:39
















2












$begingroup$

Let me try and make the question more precise. You are given a short exact sequence $0to Mto Nto Fto 0$, with $F$ flat and you want to prove that, for every $E$, the sequence $0to Motimes Eto Notimes Eto Fotimes Eto 0$ is exact.



Take an exact sequence $0to Xto Yto Eto 0$ with $Y$ projective and build the diagram with exact rows and columns
$$require{AMScd}
begin{CD}
{} @. {} @. {} @. 0 \
@. @. @. @VVV \
{} @. Motimes X @>a>> Notimes X @>b>> Fotimes X @>>> 0 \
@. @VcVV @VdVV @VeVV @. \
0 @>>> Motimes Y @>f>> Notimes Y @>g>> Fotimes Y @>>> 0 \
@. @VhVV @ViVV @VjVV @. \
{} @. Motimes E @>k>> Notimes E @>l>> Fotimes E @>>> 0 \
@. @VVV @VVV @VVV @. \
{} @. 0 @. 0 @. 0
end{CD}
$$
where we have used the fact that $F$ and $Y$ are flat.



Here's the diagram chasing. Let $xinker k$; then $x=h(y)$ and $if(y)=kh(y)=0$, so $f(y)inker i$; hence $f(y)=d(z)$. Since $eb(z)=gd(z)=gf(z)=0$, we conclude $b(z)inker e$, so $b(z)=0$. This implies $z=a(t)$.



Thus $fc(t)=da(t)=d(z)=f(y)$; since $f$ is injective, we have $y=c(t)$ and finally
$$
x=h(y)=hc(t)=0
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for graph chasing. Actually, I want a direct proof without introducing another short exact sequence.
    $endgroup$
    – Andrews
    Dec 24 '18 at 12:39














2












2








2





$begingroup$

Let me try and make the question more precise. You are given a short exact sequence $0to Mto Nto Fto 0$, with $F$ flat and you want to prove that, for every $E$, the sequence $0to Motimes Eto Notimes Eto Fotimes Eto 0$ is exact.



Take an exact sequence $0to Xto Yto Eto 0$ with $Y$ projective and build the diagram with exact rows and columns
$$require{AMScd}
begin{CD}
{} @. {} @. {} @. 0 \
@. @. @. @VVV \
{} @. Motimes X @>a>> Notimes X @>b>> Fotimes X @>>> 0 \
@. @VcVV @VdVV @VeVV @. \
0 @>>> Motimes Y @>f>> Notimes Y @>g>> Fotimes Y @>>> 0 \
@. @VhVV @ViVV @VjVV @. \
{} @. Motimes E @>k>> Notimes E @>l>> Fotimes E @>>> 0 \
@. @VVV @VVV @VVV @. \
{} @. 0 @. 0 @. 0
end{CD}
$$
where we have used the fact that $F$ and $Y$ are flat.



Here's the diagram chasing. Let $xinker k$; then $x=h(y)$ and $if(y)=kh(y)=0$, so $f(y)inker i$; hence $f(y)=d(z)$. Since $eb(z)=gd(z)=gf(z)=0$, we conclude $b(z)inker e$, so $b(z)=0$. This implies $z=a(t)$.



Thus $fc(t)=da(t)=d(z)=f(y)$; since $f$ is injective, we have $y=c(t)$ and finally
$$
x=h(y)=hc(t)=0
$$






share|cite|improve this answer









$endgroup$



Let me try and make the question more precise. You are given a short exact sequence $0to Mto Nto Fto 0$, with $F$ flat and you want to prove that, for every $E$, the sequence $0to Motimes Eto Notimes Eto Fotimes Eto 0$ is exact.



Take an exact sequence $0to Xto Yto Eto 0$ with $Y$ projective and build the diagram with exact rows and columns
$$require{AMScd}
begin{CD}
{} @. {} @. {} @. 0 \
@. @. @. @VVV \
{} @. Motimes X @>a>> Notimes X @>b>> Fotimes X @>>> 0 \
@. @VcVV @VdVV @VeVV @. \
0 @>>> Motimes Y @>f>> Notimes Y @>g>> Fotimes Y @>>> 0 \
@. @VhVV @ViVV @VjVV @. \
{} @. Motimes E @>k>> Notimes E @>l>> Fotimes E @>>> 0 \
@. @VVV @VVV @VVV @. \
{} @. 0 @. 0 @. 0
end{CD}
$$
where we have used the fact that $F$ and $Y$ are flat.



Here's the diagram chasing. Let $xinker k$; then $x=h(y)$ and $if(y)=kh(y)=0$, so $f(y)inker i$; hence $f(y)=d(z)$. Since $eb(z)=gd(z)=gf(z)=0$, we conclude $b(z)inker e$, so $b(z)=0$. This implies $z=a(t)$.



Thus $fc(t)=da(t)=d(z)=f(y)$; since $f$ is injective, we have $y=c(t)$ and finally
$$
x=h(y)=hc(t)=0
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 11:52









egregegreg

185k1486207




185k1486207












  • $begingroup$
    Thanks for graph chasing. Actually, I want a direct proof without introducing another short exact sequence.
    $endgroup$
    – Andrews
    Dec 24 '18 at 12:39


















  • $begingroup$
    Thanks for graph chasing. Actually, I want a direct proof without introducing another short exact sequence.
    $endgroup$
    – Andrews
    Dec 24 '18 at 12:39
















$begingroup$
Thanks for graph chasing. Actually, I want a direct proof without introducing another short exact sequence.
$endgroup$
– Andrews
Dec 24 '18 at 12:39




$begingroup$
Thanks for graph chasing. Actually, I want a direct proof without introducing another short exact sequence.
$endgroup$
– Andrews
Dec 24 '18 at 12:39











0












$begingroup$

We can choose a short exact sequence $0to Kto Lto Eto 0 $ s.t. $L$ is free module, and use properties: $(1),otimes $ is right exact $(2)L$ is free ($implies$flat) $(3) F$ is flat for diagram



$$Notimes Kto Motimes Kto Fotimes K$$
$$downarrow quadquadquadquaddownarrow quadquadquadquaddownarrow$$
$$Notimes Lto Motimes Lto Fotimes L$$
$$downarrow quadquadquadquaddownarrow quadquadquadquaddownarrow$$
$$Notimes Eto Motimes Eto Fotimes E$$
Snake Lemma then yields $quad0 to Notimes Eto Motimes Eto Fotimes E to0$.



Using the same idea, we can show this:



$0 to F' to F to F''$ is exact and $F''$ is flat, then $F'$ is flat $iff F$ is flat.



Especially, for exact sequence $0 to F^0to F^1 to cdots to F^n$, if $F^1, cdots, F^n$ are flat, then $F^0$ is flat.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm confused about how you're applying the snake lemma here.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 10:58










  • $begingroup$
    @AniruddhAgarwal See the other answer and try to find the snake :)
    $endgroup$
    – Andrews
    Dec 25 '18 at 10:44
















0












$begingroup$

We can choose a short exact sequence $0to Kto Lto Eto 0 $ s.t. $L$ is free module, and use properties: $(1),otimes $ is right exact $(2)L$ is free ($implies$flat) $(3) F$ is flat for diagram



$$Notimes Kto Motimes Kto Fotimes K$$
$$downarrow quadquadquadquaddownarrow quadquadquadquaddownarrow$$
$$Notimes Lto Motimes Lto Fotimes L$$
$$downarrow quadquadquadquaddownarrow quadquadquadquaddownarrow$$
$$Notimes Eto Motimes Eto Fotimes E$$
Snake Lemma then yields $quad0 to Notimes Eto Motimes Eto Fotimes E to0$.



Using the same idea, we can show this:



$0 to F' to F to F''$ is exact and $F''$ is flat, then $F'$ is flat $iff F$ is flat.



Especially, for exact sequence $0 to F^0to F^1 to cdots to F^n$, if $F^1, cdots, F^n$ are flat, then $F^0$ is flat.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm confused about how you're applying the snake lemma here.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 10:58










  • $begingroup$
    @AniruddhAgarwal See the other answer and try to find the snake :)
    $endgroup$
    – Andrews
    Dec 25 '18 at 10:44














0












0








0





$begingroup$

We can choose a short exact sequence $0to Kto Lto Eto 0 $ s.t. $L$ is free module, and use properties: $(1),otimes $ is right exact $(2)L$ is free ($implies$flat) $(3) F$ is flat for diagram



$$Notimes Kto Motimes Kto Fotimes K$$
$$downarrow quadquadquadquaddownarrow quadquadquadquaddownarrow$$
$$Notimes Lto Motimes Lto Fotimes L$$
$$downarrow quadquadquadquaddownarrow quadquadquadquaddownarrow$$
$$Notimes Eto Motimes Eto Fotimes E$$
Snake Lemma then yields $quad0 to Notimes Eto Motimes Eto Fotimes E to0$.



Using the same idea, we can show this:



$0 to F' to F to F''$ is exact and $F''$ is flat, then $F'$ is flat $iff F$ is flat.



Especially, for exact sequence $0 to F^0to F^1 to cdots to F^n$, if $F^1, cdots, F^n$ are flat, then $F^0$ is flat.






share|cite|improve this answer











$endgroup$



We can choose a short exact sequence $0to Kto Lto Eto 0 $ s.t. $L$ is free module, and use properties: $(1),otimes $ is right exact $(2)L$ is free ($implies$flat) $(3) F$ is flat for diagram



$$Notimes Kto Motimes Kto Fotimes K$$
$$downarrow quadquadquadquaddownarrow quadquadquadquaddownarrow$$
$$Notimes Lto Motimes Lto Fotimes L$$
$$downarrow quadquadquadquaddownarrow quadquadquadquaddownarrow$$
$$Notimes Eto Motimes Eto Fotimes E$$
Snake Lemma then yields $quad0 to Notimes Eto Motimes Eto Fotimes E to0$.



Using the same idea, we can show this:



$0 to F' to F to F''$ is exact and $F''$ is flat, then $F'$ is flat $iff F$ is flat.



Especially, for exact sequence $0 to F^0to F^1 to cdots to F^n$, if $F^1, cdots, F^n$ are flat, then $F^0$ is flat.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 9:12

























answered Dec 24 '18 at 9:06









AndrewsAndrews

1,2812422




1,2812422












  • $begingroup$
    I'm confused about how you're applying the snake lemma here.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 10:58










  • $begingroup$
    @AniruddhAgarwal See the other answer and try to find the snake :)
    $endgroup$
    – Andrews
    Dec 25 '18 at 10:44


















  • $begingroup$
    I'm confused about how you're applying the snake lemma here.
    $endgroup$
    – Aniruddh Agarwal
    Dec 24 '18 at 10:58










  • $begingroup$
    @AniruddhAgarwal See the other answer and try to find the snake :)
    $endgroup$
    – Andrews
    Dec 25 '18 at 10:44
















$begingroup$
I'm confused about how you're applying the snake lemma here.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 10:58




$begingroup$
I'm confused about how you're applying the snake lemma here.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 10:58












$begingroup$
@AniruddhAgarwal See the other answer and try to find the snake :)
$endgroup$
– Andrews
Dec 25 '18 at 10:44




$begingroup$
@AniruddhAgarwal See the other answer and try to find the snake :)
$endgroup$
– Andrews
Dec 25 '18 at 10:44


















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