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trivial group, since any map f: G → A into any group A for which fi: H → A is trivial (i.e. for which (1,2) is sent to 1) will identify the normal closure of H, that is the whole of G. And the kernel of f is obviously the identity id: G → G. So im(i)=ker(coker(i))= id_G. The image object is thus G, not H.
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However, i think that for the inclusion of a non-normal subgroup H in G, the image (in groups) of that morphism is the normal closure of H in G, not just H: Take as an example G=S_3 and H = {(1,2), id}. We have the inclusion morphism i: H → G. The cokernel of i is the trivial morphism G → 1 into the
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C the bijections, so that g is an image of f, but g is not equal to f as it is not even a map from A to B. I think it should read "f is monic iff f is an image of f". Similarly in the definition, it should read "an image of f is..." rather than "the image of f is...", or alternatively define an
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In the section ‘Examples’, it is stated: ″In many concrete categories such as groups, abelian groups and (left- or right) modules, the image of a morphism is the image of the correspondent morphism in the category of sets.″
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Indeed the statement that f is monic iff f = Im(f) is false - or at least "=" isn't the right relator to use. For example, in the category of sets we can have three distinct singleton sets (objects) A, B, C, with f:A-:
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Clarification added. The W in the diagram is an arbitrary object, witnessing the universality of the equalizer; this is the standard definition of
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The article doesn't really explain what the object 'I' is. Is it some special object, or just any object?
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Knowledge. If you would like to participate, please visit the project page, where you can join
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Can anybody verify this and/or check/correct the cited sentence?
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equivalence class of morphisms to be the image of f.
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