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!
Thanks in advance :)
 A: 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).
A: 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.
