3
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The problem is 'Let C be an abelian category such that all objects in C are injective. Prove that all abjects are projective.' If C has enough projectives, then the 'Ext' functor can be defined. Thus the problem is trivial as $Ext^{n}(M,N)=0$ for all objects M and N. But how can I prove that without the property 'enough projectives'?

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  • $\begingroup$ Both of these conditions are equivalent to every short exact sequence splitting ("semisimplicity"), with no appeal to the Ext functor. $\endgroup$ – Qiaochu Yuan Dec 5 '15 at 5:45
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    $\begingroup$ In any case, you can use Yoneda Ext (defined in terms of extensions), which amounts to the same thing. $\endgroup$ – Qiaochu Yuan Dec 5 '15 at 6:04

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