2

A more intuitive question: In may minmal knowledge about category, I know that any two colimits of a category commute. In the category of $A$-modules over a commutative ring $A$ with unit, the tensor product is such a colimit. But so are the image and the cokernel. But in general, we do not have $\mathrm{im}(f \otimes_A g) \cong \mathrm{im}(f) \otimes_A \mathrm{im}(g)$ or $\mathrm{coker}(f \otimes_A g) \cong \mathrm{coker}(f) \otimes_A \mathrm{coker}(g)$ for two $A$-linear maps $f$ and $g$ (e.g. here or here). What do I miss?

Gargantuar
  • 726
  • 1
    What makes you think tensor product is a colimit? What is it a colimit of? – SVG Jul 29 '23 at 20:00
  • Also to FShrike below: My question is probably ill-posed. I thought that the universal property of tensor products could be generalised as some colimit (as one usually does with with e.g. the univ. prop. of direct sums, quotients, cokernels). The problem is that the univ. prop. of tensor products involves bilinear maps, which do not lie in the category of $A$-modules anymore. – Gargantuar Jul 29 '23 at 21:32
  • @IDC Tensor product is definitely a colimit. This said, there is a difference between "each $M\otimes-$ commutes with colimits (true) and "$-\otimes-$ commutes with colimits" (false). – fosco Jul 30 '23 at 08:20
  • @fosco But this seems to be not true, e.g. $\mathrm{im}(\mathrm{id}_M \otimes_A f) \not\cong M \otimes_A \mathrm{im}(f)$. Take $\mathbb Z\to\mathbb Z$, $n\mapsto2n$ and tensor with $\mathbb Z/2$. – Gargantuar Jul 30 '23 at 08:30
  • I think the reason is that the image is not a colimit, it's a colimit of a limit (because $\text{im} = \ker \circ ,\text{coker}\cong \text{coker}\circ\ker$), and tensoring fails to preserve the limit part. But now I don't have a pencil at hand to elaborate a more precise statement... – fosco Jul 30 '23 at 08:41

1 Answers1

2

If we have two short exact sequences $$0\to A_i\xrightarrow{\iota_i}B_i\xrightarrow{\pi_i}C_i\to0,$$ $i=1, 2$, we may show that $\pi_1\otimes\pi_2\colon B_1\otimes B_2\to C_1\otimes C_2$ is surjective, and thus fits in a SES $$0\to K\xrightarrow{k}B_1\otimes B_2\xrightarrow{\pi_1\otimes\pi_2}C_1\otimes C_2\to0.$$ Consequently, the property "is a cokernel" is preserved, but the "is a cokernel of $\iota_1\otimes\iota_2$" is not guaranteed. In other words, "left adjoints preserve colimits" still works. Hope this helps :)

Laura
  • 352
  • But what exactly do I miss? Is the tensor product not a colimit? Strangely, after some thought, it seems that $\mathrm{im}$ and $\mathrm{coker}$ commute with the functor $M \otimes_A -$. So is this a colimit? – Gargantuar Jul 29 '23 at 18:57
  • I'll let someone else explain it better. I know what I want to say but I can't get it out in a satisfactory way right now, sorry :( – Laura Jul 29 '23 at 19:34
  • As I said, my knowledge about category theory is really minimal, but I think I know what you wanted to say: For the isomorphism to hold, one needs injectivity and surjectivity of the morphism. But tensor products are only right-exact, hence only preserving surjectivity. (Btw, my above comment is wrong, in general $\mathrm{im}(\mathrm{id}_M\otimes_A f) \not\cong M\otimes_A\mathrm{im}(f)$.) – Gargantuar Jul 29 '23 at 19:37
  • @Gargantuar $\newcommand{\coker}{\mathrm{coker}}$In what sense is the tensor product a colimit? I repeat the question because you should address this. If you want $\coker(f\otimes g)\cong\coker(f)\otimes\coker(g)$ then you need someway of interpreting $f\otimes g$ as a colimit. Moreover you need some way of representing this all with a product diagram to more rigorously use the Fubini theorem – FShrike Jul 29 '23 at 20:35
  • $\newcommand{\colim}{\operatorname{colim}}$To Laura, I ask a similar question. The OP was talking about the identity $\colim_{i\in I}\colim_{j\in J}D(i,j)\cong\colim_{(i,j)\in I\times J}D\cong\colim_{j\in J}\colim_{i\in I}D(i,j)$, not about left adjoints being cocontinuous. Even so, I don't think the arrow $\pi_1\otimes\pi_2$ arises from a left adjoint, because the left adjoints in question need to be $B_1\otimes(-)$ or similar (giving arrows of form $1\otimes\pi_2$, e.g.), which don't really fit here if I'm not mistaken – FShrike Jul 29 '23 at 20:39
  • Finally, "$F$ preserves colimits" is different to: "for any diagram $D$, $F(\mathrm{colim}D)$ is also a colimit (of some other diagram)" – FShrike Jul 29 '23 at 20:39
  • @FShrike if I understood correctly, the OP was worried about tensor products breaking cokernels. In that case it would be the case that cocontinuity breaks, but this does not happen because of the reason I mentioned in my post. The morphism $\pi_1\otimes\pi_2$, which "should" naively give $\text{coker}(\pi_1)\otimes\text{coker}(\pi_2)$, is the composite of $B_1\otimes\pi_2$ and $\pi_2\otimes C_2$, two instances of a tensor product functor. – Laura Jul 29 '23 at 21:19