When one studies homological algebra one learns some basic stuff-diagram chasing, long exact sequences associated to short exact sequence of complexes and so on. Usually one works out the details with the one direction of arrows (homology) and the dual results are "left to the reader" and are quite analogous. However I met one result which striked me: it is easy that each module is an image of projective module (in fact free module) however it is surprisingly not so easy to prove that each module can be embedded in an injective module despite the fact, that the definition of injective module is literally dual to projective. I would like to know more such examples, where there is some result which is true and easy but the (honestly) dual result is more difficult to prove/not analogous in proof/no longer true.
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4$\begingroup$ Projectivity and injectivity may be dual, but categories of modules don't behave well under duality - in fact the opposite of a category of modules is never even locally presentable, let alone another category of modules. $\endgroup$– Qiaochu YuanCommented Jul 6, 2016 at 22:44
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3$\begingroup$ As Qiaochu points out, the point in your example is that to prove that the category of modules (over a ring) has both enough injectives and enough projectives requires facts about modules, and isn't just about abelian categories. There are, in fact, many abelian categories that lack enough projectives or enough injectives. A typical example is the category of algebraic (aka integrable) modules for the additive group (the algebraic group $\mathbb A^1$ with operation addition). This is by definition the category of comodules for the coalgebra $\mathbb K[x]$ with comultiplication ... $\endgroup$– Theo Johnson-FreydCommented Jul 7, 2016 at 2:49
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3$\begingroup$ ... $\Delta(x) = x\otimes 1 + 1\otimes x$. This category does have enough injectives (in fact, every category of comodules over a coalgebra over a field has enough injectives) but does not have enough projectives. Its opposite category then has enough projectives and not enough injectives. What is this opposite category? It is the category of pro-finite-dimensional modules for the pro-finite-dimensional algebra $\mathbb K[[x]]$, equivalently topological modules for that power series algebra as a topological algebra. $\endgroup$– Theo Johnson-FreydCommented Jul 7, 2016 at 2:51
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3$\begingroup$ I bring up this last example (power series algebra) to give an example where you CANNOT prove that the category of modules has enough injectives, because it's just not true. So that should be an illustration that "category of modules" is a world where it's hard to prove there are injective envelopes --- you need to know things about what you're a category of modules over, and what types of modules you allow. $\endgroup$– Theo Johnson-FreydCommented Jul 7, 2016 at 2:53
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$\begingroup$ One explanation for the "non-dual" nature of the category of modules is that morita theory says that an abelian category is a category of modules over a ring precisely when it has a projective generator. This is clearly a very "left" characterization, which is not preserved by duality. In fact, we see that the class of all abelian categories with a projective generator dualizes to the class of all abelian categories with an injective generator-- exactly as in Theo's example. $\endgroup$– Phil TostesonCommented Jul 12, 2016 at 14:54
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