Does $\mathrm{Ind}(\mathcal{C})$ have enough injectives, if $\mathcal{C}$ is an abelian category?

Question

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 and the main theorem of this paper 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 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 implies this. This makes a lot more sense.

Answer 1

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 $V > U$.