# Must 2 Tensor Product Functors be Naturally Isomorphic?

Let $$R$$ be a ring, and $$M$$ an $$R$$-module. The tensor product $$(M \otimes_R -)$$ is a covariant functor from $$R$$-Mod to Ab (abelian groups).

Suppose that $$F$$ is a functor from $$R$$-Mod to Ab, which given an R-module $$N$$, associates a map $$\phi_F^N$$ along with $$F(N)$$, such that for all $$N$$, the pair $$(F(N),\phi_F^N)$$ also satisfies the universal property of the tensor product.

This means that $$F(N)$$ and $$(M \otimes_R N)$$ are isomorphic by a unique isomorphism.

What I’m wondering is if these isomorphisms are necessarily natural in $$N$$ (are the functors $$(M \otimes_R -)$$ and $$F$$ naturally isomorphic)?

I would guess not, and if so, is there a alternative category or alternative universal property that I can use and show if $$F$$ satisfies it, then I can conclude that these 2 functors are naturally isomorphic? Maybe a category of pairs $$(A, \phi)$$ of abelian groups and maps $$\phi$$ and some universal property in this category?

Please let me know if anything is not clear, and I am fairly new to all this so sorry if this is obvious!

This is not literally true as stated. For instance, suppose $$F(N)=M\otimes N$$ and $$\phi^N_F:M\times N\to F(N)=M\otimes N$$ is the usual balanced map for all $$N$$ except for one particular module $$N_0$$, and for $$N_0$$ it is instead the negative of the usual balanced map (which will still satisfy the same universal property). Then the isomorphism $$M\otimes N\to F(N)$$ obtained from $$\phi^N_F$$ will be the identity map for all $$N$$ except $$N_0$$, but for $$N_0$$ it will be the negative of the identity map. Assuming $$M\otimes N_0$$ is not $$2$$-torsion (so the negative identity is different from the identity), then these isomorphisms will not form a natural transformation (in particular, if $$N_1$$ is any module that is isomorphic but not equal to $$N_0$$, then the isomorphisms will not be natural with respect to any isomorphism $$N_1\to N_0$$).

In that particular example, $$F$$ happens to be naturally isomorphic to $$M\otimes -$$, just not via the isomorphisms given by $$\phi^N_F$$. I suspect you can also find an example where $$F$$ is not even naturally isomorphic to $$M\otimes -$$ at all, but I don't know such an example off the top of my head. For an almost-example, if you consider the functor $$\mathbb{Z}/(2)\otimes -$$ on the category of finite $$\mathbb{Z}$$-modules (rather than all $$\mathbb{Z}$$-modules), then it is objectwise isomorphic to the functor $$F(N)=\operatorname{Tor}(\mathbb{Z}/(2),N)$$, but not naturally isomorphic.

However, it is true if you make the additional assumption that the maps $$\phi_F^N$$ are natural in $$N$$ (that is, they form a natural transformation between the functors $$M\times -$$ and $$U\circ F$$ from $$R\mathtt{-Mod}$$ to $$\mathtt{Set}$$ where $$U$$ is the forgetful functor from $$\mathtt{Ab}$$ to $$\mathtt{Set}$$). You can find a high-level explanation for this in user54748's answer; here's a more hands-on verification. Let $$\alpha_N:M\otimes N\to F(N)$$ be the isomorphism induced by $$\phi^F_N$$, and suppose $$f:N\to N'$$ is a homomorphism. We wish to show that $$F(f)\circ\alpha_N=\alpha_{N'}\circ (M\otimes f).$$ To prove these homomorphisms $$M\otimes N\to F(N')$$ are equal, we can show they are equal on elements of the form $$m\otimes n$$. Note that by definition, $$\alpha_N(m\otimes n)=\phi^N_F(m,n)$$ for all $$m\in M$$ and $$n\in N$$. So, $$(F(f)\circ\alpha_N)(m\otimes n)=F(f)(\phi^N_F(m,n))$$ and $$\alpha_{N'}\circ (M\otimes f)(m\otimes n)=\phi^{N'}_F(m,f(n)).$$ But $$F(f)(\phi^N_F(m,n))=\phi^{N'}_F(m,f(n))$$ is exactly what it means for $$\phi_F$$ to be natural with respect to $$f$$, and so they are indeed equal.

This answer (as pointed out by Eric Wofsey in the comments) only works if we assume that $$φ$$ is natural in $$N$$, ie. that $$φ ∘ (\mathrm{id} × f) = Ff ∘ φ$$ for every $$f : N → N'$$. In that case the action of $$F$$ on morphisms coincides with what you get by the construction described below.

Yes, this is an important technical fact in category theory. Given a functor $$G : \mathscr D → \mathscr C$$, if there is "locally" an object $$FC ∈ \mathscr D$$ and a universal morphism $$η_C : C → GFC$$ (ie. for every $$f : C → GD$$ there exists a unique $$g : FC → D$$ such that $$Gg ∘ η = f$$) for every $$C ∈ \mathscr C$$, then $$F$$ can be pieced together into a functor defined uniquely up to a unique isomorphism, and $$F$$ is left adjoint to $$G$$.

To see how this applies in your case, notice first that the universal bilinear map $$φ_N : M × N → FN$$ can be replaced by the universal linear map $$ψ_N : N → \mathrm{Hom}_R(M, FN)$$, $$ψn := φ(-, n)$$, so that we can take $$\mathscr C = \mathrm{Mod}_R$$, $$\mathscr D = \mathrm{Mod}_R$$, and $$G = \mathrm{Hom}_R(M, -)$$.

Here's the quick proof overview of the fact in the first paragraph. First choose a universal morphism $$η_C : C → GFC$$ for every $$C$$. Now, for every $$f : C → C'$$ in $$\mathscr C$$ there is a unique morphism $$g : FC → FC'$$ such that $$Ug ∘ η = η ∘ f$$, and we can set $$g = Ff$$. Uniqueness will guarantee that $$F$$ constructed in this way is a functor, and $$η : \mathrm{Id} ⇒ GF$$ a natural transformation. Now given another choice $$η'_C : C → GF'C$$ of universal morphisms there will be a unique isomorphism $$α_C : FC → FC'$$ such that $$αη = η'$$, and this lets you prove that $$α$$ is natural. (Notice that $$α∘Ff∘η = F'f∘α∘η$$ = $$η'$$, so we have to have $$α ∘ Ff = F'F∘α$$ too.)

Technically, since you're already assuming your $$F$$ is a functor, you only need the last part of the proof, but the more general perspective is useful.

Edit: The answer assumes that $$F$$ is a functor $$\mathrm{Mod}_R → \mathrm{Mod}_R$$, since that's what you need to express the usual universal property of tensor product over a commutative $$R$$ (that $$R$$-bilinear maps $$M × N → P$$ correspond to $$R$$-linear maps $$M ⊗_R N → P$$), if you really want to consider $$M ⊗_R N$$ in $$\mathrm{Ab}$$, then you get a different universal property (that $$R$$-balanced maps $$M × N → A$$ correspond to additive maps $$M ⊗_R N → A$$), but it is again unique by the same argument (you can take $$\mathrm{Hom}(M, -)$$ for $$G : \mathrm{Ab} → \mathrm{Mod}_R$$, because $$\mathrm{Hom}(M, A)$$ has a canonical $$R$$-module structure).

• How does this apply to the question, though? What if the functor $F$ in the question happens to agree with your canonically constructed functor on objects, but does something totally different on morphisms? – Eric Wofsey Aug 8 '19 at 2:45
• For a really simple example, consider a group $G$ with an outer automorphism $F$, and consider $G$ as a 1-object category. We can consider $F$ as a functor from this category to itself, and it is objectwise isomorphic to the identity (so for each object $N$, $F(N)$ has the same universal property as $Id(N)$). But $F$ is not naturally isomorphic to the identity. – Eric Wofsey Aug 8 '19 at 2:47
• @EricWofsey: thank you, I've edited the (non)answer, and I can delete it if it would be more appropriate. I overlooked that the action of $F$ on morphisms doesn't have to "respect" $φ$. – user54748 Aug 8 '19 at 3:35
• I mean, I suspect what you wrote is what OP is looking for--they seem to be mainly looking for the correct conditions they need to get a natural isomorphism, rather than wanting the literal answer to their initial question. – Eric Wofsey Aug 8 '19 at 3:38