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How can we prove that $X$ is isomorphic to $Y$? Note: all rows and columns are exact and diagram is commutative. enter image description here

If we can do the following transformation such that $\operatorname{H}(H'‎\rightarrow‎ I'‎\rightarrow‎ J)=X$ and $\operatorname{H}(A‎\rightarrow‎ B'‎\rightarrow‎ C')=Y$ then we have $$0=\operatorname{H}(D‎\rightarrow‎ E‎\rightarrow‎ F)\rightarrow‎‎ \operatorname{H}(H'‎\rightarrow‎ I'‎\rightarrow‎ J)‎\rightarrow‎‎ \operatorname{H}(A‎\rightarrow‎ B'‎\rightarrow‎ C')‎\rightarrow \operatorname{H}(E‎\rightarrow‎ F‎\rightarrow‎ G)=0‎$$ as required.

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The first key step to the problem is to understand what the map from $X$ to $Y$ is. Do you know how to describe it and to prove that it is well-defined? If not, here's a start: take $x \in X$ and send it to an element $c \in C$. From there, you can do one of two things. Either send it to $g \in G$, but then you get $g = 0$ so that's not very helpful. So instead, you should lift it to an element $b \in B$. Of course, we may want to keep track of what happens if you lift to $b \in B$ while your best friend lifts to $b' \in B$ ... –  Michael Joyce Dec 17 '12 at 15:33
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We know that if $X=Y=0$ in this diagram then $H(D‎\rightarrow‎ E‎\rightarrow‎ F)‎\rightarrow‎ H(H‎\rightarrow‎ I ‎\rightarrow‎ J)‎\rightarrow‎ H(A‎\rightarrow‎ B‎\rightarrow‎ C)‎\rightarrow‎ H(E‎\rightarrow‎ F‎\rightarrow‎ G)$ is an exact sequence of homology modules.Now,I think that we can convert lower and upper exact rows to two complexes such that $X,Y$ are homology modules of these. –  Angel Dec 18 '12 at 20:04

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up vote 5 down vote accepted

Before I got a handle on spectral sequences, I always hated it when someone responded to my questions by saying "spectral sequences," but very soon I'm going to be guilty of the same sin.

Let's regard this as a double complex. I like cohomological conventions, so I'm going to flip your diagram upside down before regarding it as a double complex. For concreteness, let's put $A$ in degree (0,0), $B$ in degree (1,0), $E$ in degree (0,1), and so forth. If you draw it all out, you'll see that $X$ winds up in position (2, -1) and $Y$ in position (-1, 3). Let's call the double complex $S$ and the associated total complex $T$.

Now think about the associated spectral sequence where on the 0th page, we start with the vertical differentials. Now, observe that all the terms on the diagonal of total degree 1 and 2 are all 0, since in the diagram, all the vertical arrows are exact at those positions. This means that $H^1(T) = 0$ and $H^2(T) = 0$.

Now consider the other spectral sequence, where on the zeroth page, we start with the horizontal differentials. We have to go down several pages on this spectral sequence, but it's straightforward how to proceed. We see that $E_1^{2,-1} = X$, and both above it and below it are 0's, so $E_2^{2,-1} = X$ also. Now the second-page differentials also map $E_2^{2,-1} = X$ to 0's, so $E_3^{2,-1} = X$ also. Same thing with the third-page, and on the fourth page, we see that $E_4^{2,-1} = X$ and $E_4^{-1, 3} = Y$, and the differential goes from $X$ to $Y$. Let's call this differential $f$. This means on the fifth page, we get $E_5^{2,-1} = \ker f$ and $E_5^{-1,3} = \mathrm{coker}\, f$. After this, the differentials all collapse, so $E_\infty^{2,-1} = \ker f$ and $E_\infty^{-1,3} = \mathrm{coker}\, f$.

On the other hand, even on page 1, all other terms on the diagonals of total degree 1 and 2 are 0's. This tells us that $H^1(T) = \ker f$ and $H^2(T) = \mathrm{coker}\, f$.

Now, comparing against the first spectral sequence, we see that this means that $f$ is an isomorphism. I guess if you were really careful and went through the construction of the spectral sequence, you could even figure out what the isomorphism is.

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Thanks.It's true but I wanted a simple proof. –  Angel Jan 1 '13 at 9:24

In favor of this article (which should be well known to everyone in my opinion): The snake lemma by J. Wise, there is no need to take elements for diagram chasing; and the same technique applies to this diagram.

[EDIT: I finally found the time to create the diagrams.]

$X,A,B,C,D,E$ and $F$ (and $G$ via the identity map) have maps to $G$ and we can replace them by their kernels. By exactness, this doesn't effect $X$, $A,D$ and $E$ and gives a commutative diagram with exact columns and rows

where $C' = \ker(C\to G)$, $B' = \ker(B\to G)$ and $F' = \ker(F\to G)$.

Now $A$ has maps to $B', E, F'$ (and trivial maps to $I$, $C'$ and $J$). So if we take cokernels and denote these by double-primes $(\cdot)''$ we get the commutative diagram with exact columns and rows

.

Repeating the Arguments for the lower rectangle, i.e. taking kernels to $J$ and cokernels from $D$ we get a staircase

where all the non-trivial maps are isomorphisms. Hence $X$ is isomorphic to $Y$.

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,In the first diagram,Why $ 0‎\rightarrow‎ B'‎\rightarrow‎ F'‎\rightarrow‎ J‎$ is exact? –  Angel Jan 3 '13 at 19:11
    
The exactness at $F'$? Well, I don't see an element free proof of it right now (you are right, it's not trivial), but I'm sure it exists. Anyway let's say we can talk about elements and maps and give them names $f\colon B\to F$, $g\colon F\to J$ and the inclusion $i_{F'}\colon F'\to F$ as well as the restrictions (!) $\tilde{f}\colon B'\to F'$ and $\tilde{g}\colon F'\to J$. Since $\tilde{f}$ is the restriction of $f$ to $B'$, $im(\tilde{f}) = i_{F'}^{-1}(im f)$. Hence, by exactness, $im(\tilde{f}) = i_{F'}^-1(\ker g) = \ker(g i_{F'}) = \ker(\tilde{g})$ as desired. –  Ben A. Jan 4 '13 at 0:48

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