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Using Blackboard beamer theme version 0.2 I can create presentation slides that imitate the Apple’s Keynote Blackboard theme and I get to use LaTex which gives me full power to present any mathematical formula I desire!

When creating my first latex presentation slides, using the blackboard beamer theme, I found these two websites helpful.

Introduction to Beamer:
http://www.mathematik.uni-wuerzburg.de/~gerdts/beamer_intro.pdf

Making Presentations with LaTeX Guidelines:

Here is my presentation:       Back in November 2007 Anton Geraschenko had an interesting post at the Secret Blogging Seminar based on a preprint by George Bergman. His original preprint is now on the arXiv, where this is an updated version accepted by journal Theory and Applications of Categories, reformatted with their cls file. Old Lemma 12 dropped (referee noted it was immediate consequence of Lemma 11); some typos fixed, and wording cleaned up. Homological algebra is full of diagram chasing arguments.

Here is an introduction to his paper:
Diagram-chasing arguments frequently lead to “magical” relations between distant points of diagrams: exactness implications, connecting morphisms, etc.. These long connections are usually composites of short “unmagical” connections, but the latter, and the objects they join, are not visible in the proofs. I try to remedy this situation.
Given a double complex in an abelian category, we consider, for each object $A$ of the complex, the familiar horizontal and vertical homology objects at $A$, and two other objects, which we name the “donor” $A_\Box$ and and the “receptor” ${^\Box}A$ at $A$. For each arrow of the double complex, we prove the exactness of a 6-term sequence of these objects (the “Salamander Lemma”). Standard results such as the 3×3-Lemma, the Snake Lemma, and the long exact sequence of homology associated with a short exact sequence of complexes, are obtained as easy applications of this lemma.
We then obtain some generalizations of the last of the above examples, getting various exact diagrams from double complexes with all but a few rows and columns exact.
The total homology of a double complex is also examined in terms of the constructions we have introduced. We end with a brief look at the world of triple complexes, and two exercises. Diagram chasing is a method of mathematical proof used especially in homological algebra. Homological algebra studies, in particular, the homology of chain complexes in abelian categories – therefore the name. From a modern perspective, homological algebra is the study of algebraic objects, (such as groups, rings or Lie algebras, or sheaves of such objects), by ‘resolving them’, replacing them by more stable objects whose homotopy category is the derived category of an abelian category. Given a commutative diagram, a proof by diagram chasing involves the formal use of the properties of the diagram, such as injective or surjective maps, or exact sequences. A syllogism is constructed, for which the graphical display of the diagram is just a visual aid. It follows that one ends up "chasing" elements around the diagram, until the desired element or result is constructed or verified. Examples of proofs by diagram chasing include those typically given for the Snake Lemma, Four Lemma, Five Lemma, Nine Lemma, and Zig-Zag Lemma.

Definition 1 An exact sequence is a sequence, either finite or infinite, of objects and morphisms between them such that the image of one morphism equals the kernel of the next.

Definition 2 A short exact sequence is a finite sequence of objects and morphisms between them such that the image of one morphism equals the kernel of the next.

Definition 3 A long exact sequence is an infinite sequence of objects and morphisms between them such that the image of one morphism equals the kernel of the next.

Formally, an exact sequence is a sequence of maps $\displaystyle \alpha_{i}:A_{i}\rightarrow A_{i+1}$

between a sequence of spaces ${A_{i},}$ which satisfies $\displaystyle im(\alpha_{i})=\ker(\alpha_{i+1}),$

where ${im}$ denotes the image and ${\ker}$ the group kernel. That is, ${a\in A_{i},\alpha_{i}(a)=0}$ iff ${a=\alpha_{i-1}(b)}$ for some ${b\in A_{i-1}}$. It follows that ${\alpha_{i+1}\circ\alpha_{i}=0.}$ The notion of exact sequence makes sense when the spaces are groups, modules, chain complexes, or sheaves. The notation for the maps may be suppressed and the sequence written on a single line as $\displaystyle \cdots\rightarrow A_{i-1}\rightarrow A_{i}\rightarrow A_{i+1}\rightarrow\cdots.$

1. A short exact sequence: $\displaystyle 0\rightarrow A\rightarrow B\rightarrow C\rightarrow0.$

beginning and ending with zero, meaning the zero module ${\{0\}}$.

2. A long exact sequence: $\displaystyle \cdots\rightarrow A\rightarrow B\rightarrow C\rightarrow\cdots.$

Special information is conveyed when one of the spaces is the zero module. For instance, the sequence $\displaystyle 0\rightarrow A\rightarrow B$

is exact iff the map is injective. Similarly, $\displaystyle A\rightarrow B\rightarrow0$

is exact iff the map is surjective.

In homological algebra, given a short exact sequence ${0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0}$ of ${G}$-modules, there is a canonical long exact sequence $\displaystyle 0\rightarrow H^{0}(G,M')\rightarrow\cdots\rightarrow H^{i}(G,M'')\stackrel{\delta_{i}}{\rightarrow}H^{i+1}(G,M')\rightarrow \\ H^{i+1}(G,M)\rightarrow H^{i+1}(G,M'')\cdots,$

where the ${\delta_{i}}$ are certain "connecting homomorphisms" (or "snake maps"). This can be deduced from the Snake Lemma. For the proof the latter, one can engage in "diagram chasing". One can see, in the movie It’s My Turn (1980) a proof given for the Snake Lemma using diagram chasing. To define ${\delta}$: given ${x''\in\ker(\gamma)\subseteq C}$, lift ${x}$" to ${B}$, push it into ${B'}$ by ${\beta}$, then check that the image has a preimage in ${A'}$. Then verify that the result is well-defined, et cetera.

Lemma 1 (Snake Lemma) Given the commuting diagram in which the rows are exact, there is a canonical map ${\delta:\ker(\gamma)\rightarrow coker(\alpha)}$, induced by ${\delta x''=f'^{-1}\circ\beta\circ g^{-1}x''}$ such that the sequence $\displaystyle 0\rightarrow\ker(\alpha)\rightarrow\ker(\beta)\rightarrow\ker(\gamma)\stackrel{\delta}{\rightarrow} coker(\alpha)\rightarrow coker(\beta)\rightarrow \\ coker(\gamma)\rightarrow0$

is exact.

Proof: Suppose we are given a commutative diagram with exact rows. We wish to prove that the sequence $\displaystyle 0\rightarrow\ker(\alpha)\rightarrow\ker(\beta)\rightarrow\ker(\gamma)\rightarrow coker(\alpha)\rightarrow coker(\beta)\rightarrow \\ coker(\gamma)\rightarrow0$

is exact.

First we claim that if any square is commutative, then there are well-defined morphisms ${\ker(\varphi)\rightarrow\ker(\psi)}$ and ${ coker(\varphi)\rightarrow coker(\psi)}$. For example, if ${x\in\ker(\varphi)}$, then the square must commute, and so the image of ${x}$ in the top row must be in ${\ker(\psi)}$. The proof of the claim for cokernels is similar. Thus we have two sequences, $\displaystyle 0\rightarrow\ker(\alpha)\rightarrow\ker(\beta)\rightarrow\ker(\gamma){\ and\ } coker(\alpha)\rightarrow coker(\beta)\rightarrow \\ coker(\gamma)\rightarrow0,$

each of which inherits being a complex from the original diagram.

Suppose ${x\in\ker(\beta)}$ is sent to ${0\in\ker(\gamma)}$. By exactness, ${x}$ has a preimage ${x'\in A}$. Because the diagram is commutative and the bottom morphism is injective, ${y'=0}$ and so ${x'\in\ker(\alpha)}$. So the sequence $\displaystyle 0\rightarrow\ker(\alpha)\rightarrow\ker(\beta)\rightarrow\ker(\gamma)$

is exact. The proof of the claim for the cokernel sequence is similar.

So now all we need to do is find a connecting morphism ${\ker(\gamma)\rightarrow coker(\alpha)}$ such that the resulting sequence is exact at both of those points.

Suppose ${x''\in\ker(\gamma)}$. Then ${x''}$ has at least one preimage in ${B}$. So let ${x}$ and ${\widehat{x}}$ be preimages of ${x''}$. Thus ${\widehat{x}-x\mapsto0}$ and so by exactness has a preimage ${x'\in A}$. By commutativity of the diagram, ${\beta(x)}$ has a preimage ${y'}$, which is unique by injectivity of the morphism ${A'\rightarrow B'}$. But we know that the square is commutative. We wish to define ${\ker(\gamma)\rightarrow coker(\alpha)}$ by ${x''\mapsto y'+im(\alpha)}$. Observe that $\displaystyle y'+\alpha(x')\mapsto\beta(x)+\beta(\widehat{x}-x)=\beta(\widehat{x}),$

and so the choice of preimage of ${x''}$ does not affect which cokernel element we ultimately select. So now we have our connecting morphism. By applying this definition we see that $\displaystyle \ker(\beta)\rightarrow\ker(\gamma)\rightarrow coker(\alpha)\rightarrow coker(\beta)$

is a complex.

Suppose ${x''\in\ker(\gamma)}$ is sent to ${0}$ by the connecting morphism. Thus we have a diagram which is commutative. Let ${\widehat{x}}$ be the image of ${x'}$ under the morphism ${A\rightarrow B}$. Exactness of the diagram implies that ${x-\widehat{x}}$ is a preimage of ${x''}$. But ${\beta(x-\widehat{x})=0}$. So the kernel-cokernel sequence is exact at ${\ker(\gamma)}$. The proof that it is exact at ${coker(\alpha)}$ is similar.  $\Box$ 