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The OP is welcome to prove theorems about the OP's arbitrary functor that is has a naturally isomorphism to the identity functor, but I doubt anyone would find the results worthwhile. Ultimately the utility of a definition in math depends on doing something that a community of people finds interesting, and that is a matter of human judgment, not pure logic.

The OP is welcome to prove theorems about the OP's arbitrary functor, which has a natural isomorphism to the identity functor, but I doubt anyone would find the results worthwhile. Ultimately the utility of a definition in math depends on doing something that a community of people finds interesting, and that is a matter of human judgment, not pure logic.

Todd Trimble says in his answer that a problem with the OP's construction is that if you put two people in separate rooms and ask them to define a dual functor according to the OP's procedure then the two people will almost certainly not agree on the result. (EDIT: From Todd's comment below I realized that he had actually raised a different objection, where the functors themselves turn out to be different, but the point I am raising here still stands.) That isn't a fair objection, since even in situations where there is an agreed-upon natural isomorphism between two functors, it need not be the only natural isomorphism between those two functors. For example, on the category of finite-dimensional real vector spaces we have the standard natural isomorphism from the identity functor to the double dual functor, but there are many more natural isomorphisms between those two functors: for each $a \in \mathbb R^\times$ and finite-dimensional real vector space $V$ define the linear map $T_a \colon V \rightarrow V^{**}$ by $(T_a(v))(\varphi) = a\varphi(v)$ for $v \in V$ and $\varphi \in V^{*}$. Then $T_a$ is a natural isomorphism from the identity functor to the double dual functor, with $T_1$ being the standard natural isomorphism. If you put two people who only think purely logically in separate rooms and ask them to come up with a natural isomorphism between the identity and double dual functors on finite-dimensional real vector spaces then one of them might come up with $T_5$ and the other with $T_\pi$. There is no purely logical reason their results have to agree, but that doesn't mean the identity and double dual functors are not naturally isomorphic. And what I described here is not specific to vector spaces over $\mathbb R$: the same way of building extra natural isomorphisms besides a standard one works for finite-dimensional vector spaces over each field $k$ other then $\mathbb F_2$ (since $\mathbb F_2^\times = \{1\}$).

Todd Trimble says in his answer that a problem with the OP's construction is that if you put two people in separate rooms and ask them to define a dual functor according to the OP's procedure then the two people will almost certainly not agree on the result. (EDIT: From Todd's comment below I realized that he had actually raised a different objection, where the functors themselves turn out to be different, but the point I am raising here still stands.) That isn't a fair objection, since even in situations where there is an agreed-upon natural isomorphism between two functors, it need not be the only natural isomorphism between those two functors. For example, on the category of finite-dimensional real vector spaces we have the standard natural isomorphism from the identity functor to the double dual functor, but there are many more natural isomorphisms between those two functors: for each $a \in \mathbb R^\times$ and finite-dimensional real vector space $V$ define the linear map $T_a \colon V \rightarrow V^{**}$ by $(T_a(v))(\varphi) = a\varphi(v)$ for $v \in V$ and $\varphi \in V^{*}$. Then $T_a$ is a natural isomorphism from the identity functor to the double dual functor, with $T_1$ being the standard natural isomorphism. If you ask two people who only think purely logically to come up with a natural isomorphism between the identity and double dual functors on finite-dimensional real vector spaces then one of them might come up with $T_5$ and the other with $T_\pi$. There is no purely logical reason their results have to agree, but that doesn't mean the identity and double dual functors are not naturally isomorphic. And what I described here is not specific to vector spaces over $\mathbb R$: the same way of building extra natural isomorphisms besides a standard one works for finite-dimensional vector spaces over each field $k$ other then $\mathbb F_2$ (since $\mathbb F_2^\times = \{1\}$).

To get to the point quickly first, the OP has definitely constructed a natural isomorphism (with some steps missing that I fill in below.) However, it is misleading to say it is "a natural isomorphism between a vector space and its dual" because the interest in the dual space construction on finite-dimensional vector spaces is not simply forming $V^*$ from $V$ for all $V$, but also forming the dual of every linear map $f \colon V \rightarrow W$. The OP's construction has nothing to do with dual maps and that's why it is of no interest in practice. That is not a comment on logic, but on what people care about.

The OP is welcome to prove theorems about the OP's arbitrary functor that happens to be naturally isomorphic to the identity functor, but I doubt anyone would find the results worthwhile. Ultimately definitions in math depend on doing something interesting with them, and that is a matter of human judgment, not pure logic.

To get to the point quickly first, the OP has definitely constructed a natural isomorphism (with some steps missing that I fill in below.) However, it is misleading to call it "a natural isomorphism between a vector space and its dual" because the interest in the dual space construction on finite-dimensional vector spaces is not simply forming $V^*$ from $V$ for all $V$, but also forming the dual of every linear map $f \colon V \rightarrow W$. The OP's construction has nothing to do with dual maps and that's why it is of no interest in practice. That is not a comment on logic, but on what people care about.

The OP is welcome to prove theorems about the OP's arbitrary functor that is has a naturally isomorphism to the identity functor, but I doubt anyone would find the results worthwhile. Ultimately the utility of a definition in math depends on doing something that a community of people finds interesting, and that is a matter of human judgment, not pure logic.