**Question.** Let $\mathcal{C}$ be a small abelian category. Does the category $\mathrm{Ind}(\mathcal{C})$ of ind-objects of $\mathcal{C}$ have enough injectives?


I have seen many times that $\mathrm{Ind}(\mathcal{C})$ automatically has enough injectives. For example, Proposition 5 of [Akhil Mathew's note](https://amathew.wordpress.com/2012/07/25/ind-objects-in-an-abelian-category/) and the main theorem of
[this paper](https://www.ams.org/journals/tran/1972-170-00/S0002-9947-1972-0302738-0/S0002-9947-1972-0302738-0.pdf) claims such a result. The argument is something like this.


1. $\mathrm{Ind}(\mathcal{C})$ is an abelian category satisfying (AB5)
2. $``\displaystyle\bigoplus\_{X \in \mathrm{ob}\,\mathcal{C}}" X$ is a generator of $\mathrm{Ind}(\mathcal{C})$
3. Apply Theorem 1.10.1 of the Tôhoku paper


On the other hand, I found another result in [Kashiwara and Schapira's book](https://link.springer.com/chapter/10.1007/3-540-27950-4_16) which claims the exact opposite. Let $\mathrm{Mod}(k)$ be the category of $k$-vector spaces. Then Corollary 15.1.3 of this book implies that $\mathrm{Ind}(\mathrm{Mod}(k))$ does not have enough injectives.


I cannot spot any errors in both arguments. Which one is incorrect and why?




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There is a weaker claim that if $\mathcal{C}$ is an artinian abelian category, then $\mathrm{Ind}(\mathcal{C})$ has enough injectives. For example, p.356, Proposition 7 of [Gabriel's paper](http://www.numdam.org/item/?id=BSMF_1962__90__323_0) implies this. This makes a lot more sense.



Does $\mathrm{Ind}(\mathcal{C})$ have enough injectives, if $\mathcal{C}$ is an abelian category?
As Dan Petersen said in the comments, $Mod(k)$ isn't small. Note that even in this case, $Ind(Mod(k))$ only has *small* direct sums, and in particular you cannot take the direct sum that appears in point 2. of your proof sketch (I'm assuming you meant direct sum, and not tensor product); and so you don't have that generator.


In fact it's not too hard to see that there simply is no generator, and Kashiwara-Schapira prove that it also does not have enough injectives. Their proof certainly uses vector spaces of increasing dimension/cardinality (I haven't read it).


About your comment : if you restrict to some small abelian subcategory $C\subset Mod(k)$, such as "vector spaces of cardinality/dimension less than some fixed infinite cardinal", then the theorem above will apply and show that $Ind(C)$ has enough injectives.


But note that the $Ind$ construction *depends* on the universe because it is defined by freely adding *small* filtered colimits. So if you say "ok, go to a universe where $Mod(k)$ is small, and apply $Ind$ there to get something with enough injectives", that will work, but the category you obtain is not the same as the one you obtain by defining $Ind(Mod(k))$ while taking all $k$-modules of your universe of reference. Keeping track of universes, this is like $Ind^U(Mod^U(k))$ while you would be taking $Ind^V(Mod^U(k))$ for some larger universe $V > U$.


