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I have been self-studying Walter Noll's Finite-Dimensional Spaces. A tool used often that seems extremely powerful is the identification of various objects through natural isomorphisms. I have been struggling with the full concept of naturality and trying to avoid getting too involved in category theory, if possible.

One particular instance where I could not quite see an issue that was pointed out was in this Noll excerpt:

25 Tensor Products
For any linear space $\mathcal V$ there is a natural isomorphism from $\operatorname{Lin}(\Bbb R,\mathcal V)$ onto $\mathcal V$, given by $\mathbf h\mapsto\mathbf h(1)$. The inverse isomorphism associates with $\mathbf v\in\mathcal V$ the mapping $\xi\mapsto\xi\mathbf v$ in $\operatorname{Lin}(\Bbb R,\mathcal V)$. We denote this mapping by $\bf v\otimes$ (read "vee tensor") so that $$\mathbf v\otimes\xi:=\xi\mathbf v\quad\text{for all}\quad\xi\in\Bbb R.\tag{25.1}$$ In particular, there is a natural isomorphism from $\Bbb R$ onto $\Bbb R^*:=\operatorname{Lin}(\Bbb R, \Bbb R)$. It associates with every number $\eta\in\Bbb R$ the operation of multiplication with that number, so that $(25.1)$ reduces to $\eta\otimes\xi=\eta\xi$. We use this isomorphism to identify $\Bbb R^*$ with $\Bbb R$, i.e. we write $\eta=\eta\otimes$. However, when $\mathcal V\ne\Bbb R$, we do not identify $\operatorname{Lin}(\Bbb R,\mathcal V)$ with $\mathcal V$ because such an identification would conflict with the identification $\mathcal V\cong\mathcal V^{**}$ and lead to confusion.

If $\boldsymbol\lambda\in\mathcal V^*=\operatorname{Lin}(\mathcal V,\Bbb R)$, we can consider $\boldsymbol\lambda^\top\in\operatorname{Lin}(\Bbb R^*,\mathcal V^*)\cong\operatorname{Lin}(\Bbb R,\mathcal V^*)$. Using the identification $\Bbb R^*\cong\Bbb R$, it follows from $(21.3)$ and $(25.1)$ that $\boldsymbol\lambda^\top\xi=\xi\boldsymbol\lambda=\boldsymbol\lambda\otimes\xi$ for all $\xi\in\Bbb R$, i.e. that $\boldsymbol\lambda^\top=\boldsymbol\lambda\otimes$.

However, I feel like understanding this problem precisely could be enlightening. So my question is: can there be a natural isomorphism between two objects, another natural isomorphism between two other objects, and somehow a contradiction or confusion that arises because of an interrelationship?

Sticking to the Noll example would be most helpful. For example, we can make the identification $\mathcal V\cong\operatorname{Lin}(\Bbb R,\mathcal V)$ as defined by Noll's excerpt. We can also make the usual double dual identification $\mathcal V\cong\mathcal V^{**}$. Can someone point out a case where choosing to make both identifications leads to a contradiction or confusion?

Ideally, an answer that I am looking for is limited to examples involving real finite-dimensional linear spaces and relies minimally on category theoretic language.

On a side note, a suggestion of a reference to a linear algebra book that heavily motivates and makes use of natural isomorphisms would be much appreciated.

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    My first impression is that there isn't a problem, and expect that whatever actual problem they found (if any) could actually be attributed to forgetting $V^*$ and $V$ have opposite variance in $V$, obscured by the identification of $\mathbb{R}^*$ with $\mathbb{R}$. – 2017-02-10
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    Note that $V$ and $V^*$ are isomorphic, but this isomorphism is not natural, but depends on the choice of basis. $\Bbb R$ is a bit special since there is canonical basis (and really, this identification corresponds to dual base - relation $1^*(1) = 1$ implies $\eta^*(\xi) = \eta\xi$. – 2017-02-10
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    In fact, as a practical matter, rejecting $\mathcal{V} \cong \operatorname{Lin}(\mathbb{R}, \mathcal{V})$ goes against practically everything I believe in about category theory; in fact, one may get in deep enough that one would prefer to reject the idea of "element of $\mathcal{V}$" preferring to instead use "linear map $\mathbb{R} \to \mathcal{V}$" instead! You can even get a flavor of what this is like without knowing anything about category theory -- a very lowbrow example of this is when you treat elements of $\mathbb{R}^n$ as if they were $n \times 1$ matrices. – 2017-02-11
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    @Hurkyl Do you know of a way to prove the compatibility of natural isomorphisms? – 2017-06-30

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