Fix a field K and a 1-dimensional vector space $W$, and consider the functor $V\mapsto V\otimes W$ from the category of finite dimensional vector spaces to itself. In the book I am reading the author says $V$ and $V\otimes W$ are isomorphic but not naturally isomorphic. To me that means there is no natural transformation whose components are linear isomorphisms, from the identity functor to the $\otimes W$ functor.

Let me fix a non-zero vector $w\in W$, and for each finite dimensional vector space $V$ consider the map $V\to V\otimes W$ given by $v\mapsto v\otimes w$. For any linear map $f\colon V_1\to V_2$ the diagram

\begin{array} & V_1 &\longrightarrow &V_1\otimes W\\ \downarrow & & \ \ \ \ \ \downarrow \\ V_2 & {\longrightarrow} & V_2\otimes W \end{array}

commutes because a vector $v_1\in V_1$ gets sent to $f(v_1)\otimes w$ in both cases, right? Isn't this the definition of a natural transformation?

I know "natural things" have to do with morphisms being independent of choices, and my linear isomorphisms depend on a non-canonical choice of generator $w\in W$. But I can't see why, in a rigorous sense, the thing I just defined wouldn't be a natural transformation.


Putting it another way, a choice of a nonzero vector $w \in W$ is the same as a choice of isomorphism $K \rightarrow W$ ($1\in K$ gets mapped to $w$), and $V$ is naturally isomorphic to $V \otimes K$. So, yes, you are right saying that there exists a natural isomorphism for every nonzero $w \in W$.

In fact, we can prove that you've found all natural isomorphisms. Since the identity functor is represented by $K$, that is, there is an isomorphism $\operatorname{Id}_{Vect_K} \cong \operatorname{Hom}(K, -)$, we can use Yoneda lemma to get all natural transformations $\operatorname{Id}_{Vect_K} \rightarrow -\otimes W$:

$$\operatorname{Nat}(\operatorname{Id}_{Vect_K}, -\otimes W)=\operatorname{Nat}(\operatorname{Hom}(K, -), -\otimes W)=K\otimes W \cong W$$

So, each element of $W$ gives a natural transformation; for nonzero $w$ these are isomorphisms.

I suppose the author means that there is no single preffered natural isomorphism - there are many of them, all as good as others.

  • $\begingroup$ It's as if it is not natural in $W$ where we view $W$ as ranging over the full subcategory of 1-dimensional vector spaces (over $K$)... $\endgroup$ – Derek Elkins Jun 7 '18 at 22:07
  • $\begingroup$ so maybe "naturally isomorphic" means there exists a $\textit{unique}$ natural transformation (consisting of isomorphisms)? $\endgroup$ – Marco Flores Jun 10 '18 at 9:20

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