Let us define a short exact sequence 0 --> U --> V --> W --> 0 by saying that U --> V is the kernel of V --> W ("left exactness") and V --> W is the cokernel of U --> V ("right exactness"). If so, your formulation becomes is trivial in full generality: The kernel of a morphism X --> Y (in any pointed category with zero object) is the pullback of X --> Y and 0 --> Y. Therefore: (a) D' is the pullback of G --> C and 0 --> C; (b) the kernel of D --> G --> C is the pullback of D --> G --> C and 0 --> C; (c) as follows from (a) and (b), the kernel of D --> G --> C is the pullback of D --> G and D' --> C; (d) similarly, the kernel of D' --> G --> C' is the same pullback. Since this is true in every category with zero object (you don't need any normality argument), the dual statement is also always true (and everything with pushouts and cokernels instead of pullbacks and kernels is equally clear). Moreover (obviously), for the coincidence of the kernels we need only the "left exactness", and dually, for the coincidence of the cokernels we need only the "right exactness" of our original sequences. However, what Jim's says is not true in every pointed Barr exact category with cokernels: it will be true if and only if every regular epimorphism is normal (that is, is the cokernel of something, or, equivalently, the cokernel of its kernel). George -------------------------------------------------- From: "Michael Barr" <barr@math.mcgill.ca> Sent: Friday, August 26, 2011 1:03 AM To: "Categories list" <categories@mta.ca> Subject: categories: Goursat's lemma
This is something that Lambek used for diagram chasing in his book on Rings and Modules. I have discovered a very simple proof of a slight generalization.
The lemma states (as least Jim stated it this way) that if G is a submodule of A x B and C = image of G --> A, C' the image of G --> B, D the kernel of G --> B and D' the kernel of G --> A, then C/D is isomorphic to C'/D'.
Forget that G is a subgroup of A x B and just suppose you have two exact sequences 0 --> D' --> G --> C --> 0 and 0 --> D --> G --> C' --> 0, then the composites D --> G --> C and D' --> G --> C' have the same cokernel. The dual claim is that they have the same kernel. But thinking of D and D' are submodules of G, then the kernels are quite obviously D \cap D' and the original conclusion follows by duality.
However, the same thing is true for groups (both D and D', being kernels, are normal in G and hence in C, C'). I see no such simple duality argument there.
Michael
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