How to write diagram chases - Revision history

Anton: /* Mittag-Leffler Theorem */

2012-08-10T20:02:13Z

Mittag-Leffler Theorem

← Older revision		Revision as of 12:02, 10 August 2012
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	</pre>		</pre>

	== Mittag-Leffler Theorem ==		== Mittag-Leffler (?) Theorem ==

	'''~~Mittag-Leffler~~ Theorem.''' Suppose $A_\bullet$, $B_\bullet$, and $C_\bullet$ are inverse systems of abelian groups and $A_\bullet\xrightarrow{f_\bullet} B_\bullet\xrightarrow{g_\bullet} C_\bullet \to 0$ is a right exact sequence of inverse systems. If $A_\bullet$ satisfies the [http://en.wikipedia.org/wiki/Inverse_limit#Mittag-Leffler_condition Mittag-Leffler condition] (e.g. if all the maps in $A_\bullet$ are surjective), then the induced sequence $\varprojlim A_\bullet \to \varprojlim B_\bullet \to \varprojlim C_\bullet \to 0$ is exact.		'''Theorem.''' Suppose $A_\bullet$, $B_\bullet$, and $C_\bullet$ are inverse systems of abelian groups and $A_\bullet\xrightarrow{f_\bullet} B_\bullet\xrightarrow{g_\bullet} C_\bullet \to 0$ is a right exact sequence of inverse systems. If $A_\bullet$ satisfies the [http://en.wikipedia.org/wiki/Inverse_limit#Mittag-Leffler_condition Mittag-Leffler condition] (e.g. if all the maps in $A_\bullet$ are surjective), then the induced sequence $\varprojlim A_\bullet \to \varprojlim B_\bullet \to \varprojlim C_\bullet \to 0$ is exact.

	Note: $\varprojlim$ is a right adjoint functor, so it is automatically left exact.		Note: $\varprojlim$ is a right adjoint functor, so it is automatically left exact.

Anton: /* Mittag-Leffler Theorem */

2012-07-15T19:10:30Z

Mittag-Leffler Theorem

← Older revision		Revision as of 11:10, 15 July 2012
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	Note: $\varprojlim$ is a right adjoint functor, so it is automatically left exact.		Note: $\varprojlim$ is a right adjoint functor, so it is automatically left exact.

	''Proof.'' After replacing $A_n$ by the intersection of the images of the $A_N$, we may assume all the maps in the inverse system $A_\bullet$ are surjective. It suffices to show that ~~for~~ any compatible system $\{c_n\}$, representing an element of $\varprojlim C_\bullet$, is expressible as the image of some compatible system $\{b_n\}$, representing an element of $\varprojlim B_\bullet$. We construct $\{b_n\}$ by induction. Since $g_0$ is surjective, we have that $c_0 = g_0(b_0)$ for some $b_0\in B_0$. For the induction step, we chase the following diagram.		''Proof.'' After replacing $A_n$ by the intersection of the images of the $A_N$, we may assume all the maps in the inverse system $A_\bullet$ are surjective. It suffices to show that any compatible system $\{c_n\}$, representing an element of $\varprojlim C_\bullet$, is expressible as the image of some compatible system $\{b_n\}$, representing an element of $\varprojlim B_\bullet$. We construct $\{b_n\}$ by induction. Since $g_0$ is surjective, we have that $c_0 = g_0(b_0)$ for some $b_0\in B_0$. For the induction step, we chase the following diagram.

	[[File:Mittag-LefflerChase.jpg]]		[[File:Mittag-LefflerChase.jpg]]

Anton: /* $\LaTeX$ source */

2012-07-15T19:05:15Z

$\LaTeX$ source

← Older revision		Revision as of 11:05, 15 July 2012
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	\newcommand{\cok}{\mathrm{cok}\,}		\newcommand{\cok}{\mathrm{cok}\,}

	\begin{document}		\begin{document}
	~~\newcommand{\cok}{\mathrm{cok}\,~~}
	\[\xymatrix{		\[\xymatrix{
	& \ker f \ar[r]\ar[d] & \ker g \ar[r]\ar[d] & \ker h \ar[d]\\		& \ker f \ar[r]\ar[d] & \ker g \ar[r]\ar[d] & \ker h \ar[d]\\
	& \bullet \ar[r]\ar[d]_(.6)f\|(.3)\hole		& \bullet \ar[r]\ar[d]_(.6)f\|(.3)\hole
	& \bullet \ar[r]\ar[d]_(.6)g\|(.3)\hole		& \bullet \ar[r]\ar[d]_(.6)g\|(.3)\hole
	& \bullet \ar[r]\ar[d]_(.6)h\|(.3)\hole="hole" & 0 \\		& \bullet \ar[r]\ar[d]_(.6)h\|(.3)\hole="hole" & 0 \\
	0\ar[r] & \bullet \ar[r]\ar[d] & \bullet \ar[r]\ar[d] & \bullet \ar[d] \\		0\ar[r] & \bullet \ar[r]\ar[d] & \bullet \ar[r]\ar[d] & \bullet \ar[d] \\
	& \cok f \ar[r]		& \cok f \ar[r]
	\ar@{<-} `l [lu]-(.5,0) `"hole" `[rrruu]+(.5,0) `[rruuu] [rruuu]		\ar@{<-} `l [lu]-(.5,0) `"hole" `[rrruu]+(.5,0) `[rruuu] [rruuu]
	& \cok g \ar[r] & \cok h \\		& \cok g \ar[r] & \cok h \\
	%%%%		%%%%
	& & & *+[o][F-]{x} \ar@{\|->}[d]\\		& & & *+[o][F-]{x} \ar@{\|->}[d]\\
	& { }\POS p+(-.5,.5) *+{t}="t"		& { }\POS p+(-.5,.5) *+{t}="t"
	& z\ar@{\|->}[r]\ar@{\|->}[d]		& z\ar@{\|->}[r]\ar@{\|->}[d]
	\POS p+(-.5,.5) *+{z-z'}="z-zp" \ar@{\|.>}[d]+(-.5,.7)		\POS p+(-.5,.5) *+{z-z'}="z-zp" \ar@{\|.>}[d]+(-.5,.7)
	\POS p+(.5,-.5) *+{z'}="zp" \ar@{\|->}[r]		\POS p+(.5,-.5) *+{z'}="zp" \ar@{\|->}[r]
	& y \ar@{\|.>}[r] \ar@{\|.>}[d]		& y \ar@{\|.>}[r] \ar@{\|.>}[d]
	\POS p+(-.5,.5) *+{0}="0" & 0\\		\POS p+(-.5,.5) *+{0}="0" & 0\\
	& r\ar@{\|->}[r]\ar@{\|->}[d]		& r\ar@{\|->}[r]\ar@{\|->}[d]
	\POS p+(-.5,.5) *+{r-r'}="r-rp"		\POS p+(-.5,.5) *+{r-r'}="r-rp"
	\POS p+(.5,-.5) *+{r'}="rp" \ar@{\|.>}[d]		\POS p+(.5,-.5) *+{r'}="rp" \ar@{\|.>}[d]
	& w\ar@{\|.>}[r]\ar@{\|.>}[d]		& w\ar@{\|.>}[r]\ar@{\|.>}[d]
	\POS p+(.5,-.5) *+{w'}="wp" \ar@{\|.>}[r]\ar@{\|.>}[d]		\POS p+(.5,-.5) *+{w'}="wp" \ar@{\|.>}[r]\ar@{\|.>}[d]
	\POS p+(-.7,.5) *+{w-w'}="w-wp" & 0 \\		\POS p+(-.7,.5) *+{w-w'}="w-wp" & 0 \\
	& \llap{$\delta(x):=\;$} s\ar@{\|.>}[r]		& \llap{$\delta(x):=\;$} s\ar@{\|.>}[r]
	\POS p+(-.5,.5) *+{0}="0p" & 0 \\		\POS p+(-.5,.5) *+{0}="0p" & 0 \\
	\ar@{\|->} "t";"z-zp"		\ar@{\|->} "t";"z-zp"
	\ar@{\|.>} "t";"r-rp"		\ar@{\|.>} "t";"r-rp"

Anton at 04:54, 15 July 2012

2012-07-15T04:54:15Z

← Older revision		Revision as of 20:54, 14 July 2012
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	~~Diagram~~ chasing arguments are ~~supposed to be~~ impossible to write in such a way that it takes somebody less effort to read than to reproduce the argument for themselves. Many years ago, I cooked up a solution to this problem. In fact, this problem was what motivated me to learn more about the [[Salamander lemma]].		It's been said that diagram chasing arguments are impossible to write in such a way that it takes somebody less effort to read than to reproduce the argument for themselves. Many years ago, I cooked up what I think is a reasonably good solution to this problem. In fact, this problem was what motivated me to learn more about the [[Salamander lemma]] in the first place.

	= The Rules =		= The Rules =

Anton at 04:51, 15 July 2012

2012-07-15T04:51:18Z

← Older revision		Revision as of 20:51, 14 July 2012
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	= Examples =		= Examples =

			Here are a couple of examples to illustrate.

	== Connecting homomorphism in the Snake Lemma ==		== Connecting homomorphism in the Snake Lemma ==
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	\end{document}		\end{document}
	</pre>		</pre>

			[[Category:Blog]]

Anton at 04:50, 15 July 2012

2012-07-15T04:50:00Z

← Older revision		Revision as of 20:50, 14 July 2012
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	<ol start="0">		<ol start="0">
	<li> The "starting element" should be visually distinguished in some way (I like to put a circle around it). Moreover, any time you introduce a new element which is not the image or in the preimage of any other element of the diagram, it should be visually distinguished.		<li> The "starting element" should be visually distinguished in some way (I like to put a circle around it). Moreover, any time you introduce a new element which is not the image or in the preimage of any other element of the diagram, it should be visually distinguished (relax this rule as you see fit).
	<li> Two elements are connected by a '''solid arrow''' if one is '''defined''' in terms of the other using that arrow. For example, one element might be defined as the image of another under some map, or as some element of the preimage under some map.		<li> Two elements are connected by a '''solid arrow''' if one is '''defined''' in terms of the other using that arrow. For example, one element might be defined as the image of another under some map, or as some element of the preimage under some map.
	<li> Two elements are connected by a '''dotted arrow''' to indicate a relationship '''forced''' by some aspect of the problem. The relationship may be imposed by commutativity of the diagram, exactness of some row, additivity of the maps, or some other fact that was taken as input to the problem.		<li> Two elements are connected by a '''dotted arrow''' to indicate a relationship '''forced''' by some aspect of the problem. The relationship may be imposed by commutativity of the diagram, exactness of some row, additivity of the maps, or some other fact that was taken as input to the problem.
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	The point of these rules is that the graph of solid arrows must be acyclic, so any element has a unique path of solid arrows connecting it to one of the roots (the "starting elements"). This means that a reader can unambiguously figure out how each element of the diagram was constructed.		The point of these rules is that the graph of solid arrows must be acyclic, so any element has a unique path of solid arrows connecting it to one of the roots (the "starting elements"). This means that a reader can unambiguously figure out how each element of the diagram was constructed.

	= Examples~~, including how to typeset with xymatrix~~ =		= Examples =

	== Connecting homomorphism in the Snake Lemma ==		== Connecting homomorphism in the Snake Lemma ==

	The Snake Lemma says that if the two middle horizontal rows in the following commutative diagram are exact, then there is a connecting homomorphism $\def\cok{\mathrm{cok\,}} \ker h\to\cok f$ so that the 6-term sequence $\ker f\to\ker g\to\ker h\to\cok f\to\cok g\to\cok h$ is exact. Here we produce this connecting homomorphism.		The Snake Lemma says that if the two middle horizontal rows in the following commutative diagram are exact, then there is a connecting homomorphism $\def\cok{\mathrm{cok\,}} \ker h\xrightarrow\delta\cok f$ so that the 6-term sequence $\ker f\to\ker g\to\ker h\xrightarrow\delta\cok f\to\cok g\to\cok h$ is exact. Here we produce this connecting homomorphism.

	[[File:SnakeLemmaChase.jpg]]		[[File:SnakeLemmaChase.jpg]]

	=== ~~Source~~ ===		Given an element $x$ of $\ker h$, let $y$ be its image. Let $z$ be an element mapping to $y$. Let $w$ be the image of $z$. Since the right vertical column is exact, $y$ maps to zero, and since the diagram commutes, $w$ must also map to zero, so it must be the image of some $r$. Let $s$ be the image of $r$. We define $\delta(x)=s$.

			To see that this definition is independent of the choice of $z$, suppose $z'$ is some other element mapping to $y$, and define $w'$ and $r'$ accordingly. Since $z-z'$ maps to $y-y=0$, it must be the image of some element $t$. By commutativity of the diagram (and the fact that the bottom middle row is exact), $t$ must map to $r-r'$. Since the left column is exact, we have that $r-r'$ must map to zero, so $r'$ must have the same image as $r$.

			=== $\LaTeX$ source ===

	<pre>		<pre>
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	\begin{document}		\begin{document}
			\newcommand{\cok}{\mathrm{cok}\,}
	\[\xymatrix{		\[\xymatrix{
	& \ker f \ar[r]\ar[d] & \ker g \ar[r]\ar[d] & \ker h \ar[d]\\		& \ker f \ar[r]\ar[d] & \ker g \ar[r]\ar[d] & \ker h \ar[d]\\
	& \bullet \ar[r]\ar[d]_(.6)f\|(.3)\hole		& \bullet \ar[r]\ar[d]_(.6)f\|(.3)\hole
	& \bullet \ar[r]\ar[d]_(.6)g\|(.3)\hole		& \bullet \ar[r]\ar[d]_(.6)g\|(.3)\hole
	& \bullet \ar[r]\ar[d]_(.6)h\|(.3)\hole="x" & 0 \\		& \bullet \ar[r]\ar[d]_(.6)h\|(.3)\hole="hole" & 0 \\
	0\ar[r] & \bullet \ar[r]\ar[d] & \bullet \ar[r]\ar[d] & \bullet \ar[d] \\		0\ar[r] & \bullet \ar[r]\ar[d] & \bullet \ar[r]\ar[d] & \bullet \ar[d] \\
	& \cok f \ar[r]		& \cok f \ar[r]
	\ar@{<-} `l [lu]-(.5,0) `"x" `[rrruu]+(.5,0) `[rruuu] [rruuu]		\ar@{<-} `l [lu]-(.5,0) `"hole" `[rrruu]+(.5,0) `[rruuu] [rruuu]
	& \cok g \ar[r] & \cok h \\		& \cok g \ar[r] & \cok h \\
	%%%%		%%%%
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	\POS p+(.5,-.5) *+{w'}="wp" \ar@{\|.>}[r]\ar@{\|.>}[d]		\POS p+(.5,-.5) *+{w'}="wp" \ar@{\|.>}[r]\ar@{\|.>}[d]
	\POS p+(-.7,.5) *+{w-w'}="w-wp" & 0 \\		\POS p+(-.7,.5) *+{w-w'}="w-wp" & 0 \\
	& s\ar@{\|.>}[r]		& \llap{$\delta(x):=\;$} s\ar@{\|.>}[r]
	\POS p+(-.5,.5) *+{0}="0p" & 0 \\		\POS p+(-.5,.5) *+{0}="0p" & 0 \\
	\ar@{\|->} "t";"z-zp"		\ar@{\|->} "t";"z-zp"
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	''Proof.'' After replacing $A_n$ by the intersection of the images of the $A_N$, we may assume all the maps in the inverse system $A_\bullet$ are surjective. It suffices to show that for any compatible system $\{c_n\}$, representing an element of $\varprojlim C_\bullet$, is expressible as the image of some compatible system $\{b_n\}$, representing an element of $\varprojlim B_\bullet$. We construct $\{b_n\}$ by induction. Since $g_0$ is surjective, we have that $c_0 = g_0(b_0)$ for some $b_0\in B_0$. For the induction step, we chase the following diagram.		''Proof.'' After replacing $A_n$ by the intersection of the images of the $A_N$, we may assume all the maps in the inverse system $A_\bullet$ are surjective. It suffices to show that for any compatible system $\{c_n\}$, representing an element of $\varprojlim C_\bullet$, is expressible as the image of some compatible system $\{b_n\}$, representing an element of $\varprojlim B_\bullet$. We construct $\{b_n\}$ by induction. Since $g_0$ is surjective, we have that $c_0 = g_0(b_0)$ for some $b_0\in B_0$. For the induction step, we chase the following diagram.

			[[File:Mittag-LefflerChase.jpg]]

			Since $g_n$ is surjective, there is some $x$ mapping to $c_n$. Let $y=\beta_n(x)$. By commutativity of the diagram, $g_{n-1}(y)=c_{n-1}$, so $g_{n-1}(b_{n-1}-y)=c_{n-1}-c_{n-1}=0$, so by exactness of the bottom row, there is some $z$ so that $f_{n-1}(z)=b_{n-1}-y$. Since $\alpha_n$ is surjective, there is some $w$ so that $\alpha_n(w)=z$. Let $r=f_n(w)$. By commutativity of the diagram, we have $\beta_n(r)=b_{n-1}-y$, so $\beta_n(x+r)=y+b_{n-1}-y$. Moreover, we have that $g_n(x+r)=0+c_n$. So we define $b_n$ to be $x+r$.

			=== $\LaTeX$ source ===

			<pre>

			\documentclass{article}
			\usepackage[all]{xy}
			\SelectTips{cm}{10} % Use the arrowheads that agree with inline arrows.
			\everyxy={<2.5em,0em>:} % Set the unit length in xy diagrams.

			\begin{document}
			\[\xymatrix@C+1pc{
			A_{n+1}\ar[d]^{\alpha_n}\ar[r]^-{f_n}
			& B_{n+1}\ar[d]^{\beta_n}\ar[r]^{g_n}
			& C_{n+1}\ar[d]^{\gamma_n}\ar[r] & 0\\
			A_n \ar[r]^{f_{n-1}}
			& B_n \ar[r]^{g_{n-1}}
			& C_n\ar[r] & 0\\
			%%%%
			{ }\POS p-(.8,.5)*+{w}="w"
			& \llap{$b_n:=\;$}x+r \ar@{\|.>}[d] \ar@{\|.>}[r]
			\POS p-(.8,.5)*+{r}="r" \ar@{<-\|}"w"
			\POS p+(.8,.5)*+{x}="x" \ar@/^/@{\|->}[r]
			& *+[o][F-]{c_n}\ar@{\|.>}[d]\\
			{ }\POS p-(.8,.5)*+{z}="z" \ar@{<-\|}"w"
			& b_{n-1}\ar@{\|.>}[r]
			\POS p-(.8,.5)*+{b_{n-1}-y}="ed" \ar@{<-\|}"z"
			\POS p+(.8,.5)*+{y}="y"\ar@/^/@{\|.>}[r]
			& c_{n-1}
			\POS p-(.8,.5)*+{0}="zero" \ar@{<.\|}"ed"
			\ar@{\|->} "x";"y"
			\ar@{\|.>} "r";"ed"
			}\]
			\end{document}
			</pre>

Anton at 04:24, 15 July 2012

2012-07-15T04:24:28Z

@@ Line 5: / Line 5: @@
 Draw the diagram of sets which you're going to chase. The elements being chased will be drawn in a "parallel" diagram, where their relative positioning indicates which set they belong to (see the examples below). Then draw elements connected by arrows using the following rules:
-. The "starting element" should be visually distinguished in some way (I like to put a circle around it)
+<ol start="0">
-= Examples =
+Basically, whenever a new element is introduced into the diagram, connect it to an existing element using a solid arrow. Whenever you wish to connect two existing elements of your diagram, use a dotted arrow.
-= How to typeset with xymatrix =
+The point of these rules is that the graph of solid arrows must be acyclic, so any element has a unique path of solid arrows connecting it to one of the roots (the "starting elements"). This means that a reader can unambiguously figure out how each element of the diagram was constructed.

Anton: Created page with "Diagram chasing arguments are supposed to be impossible to write in such a way that it takes somebody less effort to read than to reproduce the argument for themselves. Many year..."

2012-07-15T02:06:31Z

Created page with "Diagram chasing arguments are supposed to be impossible to write in such a way that it takes somebody less effort to read than to reproduce the argument for themselves. Many year..."

New page

Diagram chasing arguments are supposed to be impossible to write in such a way that it takes somebody less effort to read than to reproduce the argument for themselves. Many years ago, I cooked up a solution to this problem. In fact, this problem was what motivated me to learn more about the [[Salamander lemma]].

= The Rules =

Draw the diagram of sets which you're going to chase. The elements being chased will be drawn in a "parallel" diagram, where their relative positioning indicates which set they belong to (see the examples below). Then draw elements connected by arrows using the following rules:

0. The "starting element" should be visually distinguished in some way (I like to put a circle around it)

= Examples =

= How to typeset with xymatrix =