3 added 20 characters in body

Regarding the (widely considered to be false) statement that "There is a canonical isomorphism between a finite-dimensional vector space V and its dual" in Reid Barton's answer, I think that the situation is a bit more interesting than that. It is a good illustration of the idea that an object may be defined to be "canonical" if it is constructed without making any choices, and the interesting point here is that there are various degrees of (in-)tolerance to choices. If we work with vector spaces of fixed finite dimension, then an isomorphism $i_E:E\to E^*$ between a vector space $E$ and its dual $E^*$ may be called canonical if

1. it does not depend on the choice of a basis for $E$ but we need a basis to define it, or
2. it does not depend on the choice of a basis and may be defined without choosing a basis, or
3. it does not depend on basis choices as above and does not even depend on $E$, in the sense that whenever $u:E\to F$ is an isomorphism between dim. vector spaces of the same finite dimension, then $u^*\circ i_F\circ u=i_E$ where $u^*$ is the transpose of $u$. This third notion of canonicity is essentially functoriality.

Concretely, given a basis $B=\{e_j\}$ of $E$ with dual basis $B^*=\{e_j^*\}$, we can construct an isomorphism $i=i_{E,B}:E\to E^*$ that maps $e_j$ to $e_j^*$. This map does not depend on the choice of basis if and only if $u^*\circ i\circ u=i$ for all $u\in \text{GL}(E)$. It is easily seen that this is equivalent to the fact that $\text{GL}_n(k)=\text{O}_n(k)$, where $k$ is the base field, $n$ is the dimension, and $\text{O}_n(k)$ is the orthogonal group of the standard (sum of squares) quadratic form. Exercise: this equality holds if and only if $n=1$ and $k$ has at most $3$ elements. Thus for $n=1$ and $\text{card}(k)\leq 3$ the map $i_{E,B}:E\to E^*$ does not depend on $B$, so we may write it simply $i_E$. When $\text{card}(k)=2$ this is not so surprising because any two one-dimensional vector spaces over the field with two elements are uniquely (hence canonically in whichever sense you like) isomorphic, but for $\text{card}(k)=3$ this is a bit more exotic. Having reached this point, we might think that we are in the funny notion 1 of canonicity (and this is what I thought some minutes ago). But in fact, still assuming that $n=1$ and $\text{card}(k)\leq 3$, we can exhibit an isomorphism $i:E\to E^*$ without any reference to a basis. Namely, define $i(0)=0$ and if $x\in E$ is nonzero, then it is a basis of $E$, and we can define $i(x)=x^*$, the only element of the dual basis. The point is that since $a^2=1$ for all nonzero scalars in $k$, this map $i$ is linear.

Conclusion: if $n=1$ and $\text{card}(k)\leq 3$ there is an isomorphism $i:E\to E^*$ that is constructed without a choice of basis, and it is functorial for isomorphisms of one-dimensional vector spaces. If $n\ge 2$ or $\text{card}(k)\ge 4$, there the map $i_{E,B}:E\to E^*$ is no such basis-choice-free isomorphismnot independent of the basis $B$.

I would guess that it is possible to find examples of phenomena like 1 above.

Regarding the (widely considered to be false) statement that "There is a canonical isomorphism between a finite-dimensional vector space V and its dual" in Reid Barton's answer, I think that the situation is a bit more interesting than that. It is a good illustration of the idea that an object may be defined to be "canonical" if it is constructed without making any choices, and the interesting point here is that there are various degrees of (in-)tolerance to choices. If we work with vector spaces of fixed finite dimension, then an isomorphism $i_E:E\to E^*$ between a vector space $E$ and its dual $E^*$ may be called canonical if

1. it does not depend on the choice of a basis for $E$ but we need a basis to define it, or
2. it does not depend on the choice of a basis and may be defined without choosing a basis, or
3. it does not depend on basis choices as above and does not even depend on $E$, in the sense that whenever $u:E\to F$ is an isomorphism between dim. vector spaces of the same finite dimension, then $u^*\circ i_F\circ u=i_E$ where $u^*$ is the transpose of $u$. This third notion of canonicity is essentially functoriality.

Concretely, given a basis $B=\{e_j\}$ of $E$ with dual basis $B^*=\{e_j^*\}$, we can construct an isomorphism $i=i_{E,B}:E\to E^*$ that maps $e_j$ to $e_j^*$. This map does not depend on the choice of basis if and only if $u^*\circ i\circ u=i$ for all $u\in \text{GL}(E)$. It is easily seen that this is equivalent to the fact that $\text{GL}_n(k)=\text{O}_n(k)$, where $k$ is the base field, $n$ is the dimension, and $\text{O}_n(k)$ is the orthogonal group of the standard (sum of squares) quadratic form. Exercise: this equality holds if and only if $n=1$ and $k$ has at most $3$ elements. Thus for $n=1$ and $\text{card}(k)\leq 3$ the map $i_{E,B}:E\to E^*$ does not depend on $B$, so we may write it simply $i_E$. When $\text{card}(k)=2$ this is not so surprising because any two one-dimensional vector spaces over the field with two elements are uniquely (hence canonically in whichever sense you like) isomorphic, but for $\text{card}(k)=3$ this is a bit more exotic. Having reached this point, we might think that we are in the funny notion 2 1 of canonicity (and this is what I thought some minutes ago). But in fact, still assuming that $n=1$ and $\text{card}(k)\leq 3$, we can exhibit an isomorphism $i:E\to E^*$ without any reference to a basis. Namely, define $i(0)=0$ and if $x\in E$ is nonzero, then it is a basis of $E$, and we can define $i(x)=x^*$, the only element of the dual basis. The point is that since $a^2=1$ for all nonzero scalars in $k$, this map $i$ is linear.

Conclusion: if $n=1$ and $\text{card}(k)\leq 3$ there is an isomorphism $i:E\to E^*$ that is constructed without a choice of basis, and it is functorial for isomorphisms of one-dimensional vector spaces. If $n\ge 2$ or $\text{k}\ge \text{card}(k)\ge 4$, there is no such basis-choice-free isomorphism.

I would guess that it is possible to find examples of phenomena like 21 above.

 
 
 
 
Regarding the (widely considered to be false) statement that "There is a canonical isomorphism between a finite-dimensional vector space V and its dual" in Reid Barton's answer, I think that the situation is a bit more interesting than that. It is a good illustration of the idea that an object may be defined to be "canonical" if it is constructed without making any choices, and the interesting point here is that there are various degrees of (in-)tolerance to choices. If we work with vector spaces of fixed finite dimension, then an isomorphism $i_E:E\to E^*$ between a vector space $E$ and its dual $E^*$ may be called canonical if
1. it does not depend on the choice of a basis for $E$, or
3. it does not depend on basis choices as above and does not even depend on $E$, in the sense that whenever $u:E\to F$ is an isomorphism between dim. vector spaces of the same finite dimension, then $u^*\circ i_F\circ u=i_E$ where $u^*$ is the transpose of $u$. This third notion of canonicity is essentially functoriality.
Concretely, given a basis $B=\{e_j\}$ of $E$ with dual basis $B^*=\{e_j^*\}$, we can construct an isomorphism $i=i_{E,B}:E\to E^*$ that maps $e_j$ to $e_j^*$. This map does not depend on the choice of basis if and only if $u^*\circ i\circ u=i$ for all $u\in \text{GL}(E)$. It is easily seen that this is equivalent to the fact that $\text{GL}_n(k)=\text{O}_n(k)$, where $k$ is the base field, $n$ is the dimension, and $\text{O}_n(k)$ is the orthogonal group of the standard (sum of squares) quadratic form. Exercise: this equality holds if and only if $n=1$ and $k$ has at most $3$ elements. Thus for $n=1$ and $\text{card}(k)\leq 3$ the map $i_{E,B}:E\to E^*$ does not depend on $B$, so we may write it simply $i_E$. When $\text{card}(k)=2$ this is not so surprising because any two one-dimensional vector spaces over the field with two elements are uniquely (hence canonically in whichever sense you like) isomorphic, but for $\text{card}(k)=3$ this is a bit more exotic. Having reached this point, we might think that we are in the funny notion 2 of canonicity (and this is what I thought some minutes ago). But in fact, still assuming that $n=1$ and $\text{card}(k)\leq 3$, we can exhibit an isomorphism $i:E\to E^*$ without any reference to a basis. Namely, define $i(0)=0$ and if $x\in E$ is nonzero, then it is a basis of $E$, and we can define $i(x)=x^*$, the only element of the dual basis. The point is that since $a^2=1$ for all nonzero scalars in $k$, this map $i$ is linear.
Conclusion: if $n=1$ and $\text{card}(k)\leq 3$ there is an isomorphism $i:E\to E^*$ that is constructed without a choice of basis, and it is functorial for isomorphisms of one-dimensional vector spaces. If $n\ge 2$ or $\text{k}\ge 4$, there is no such basis-choice-free isomorphism.
I would guess that it is possible to find examples of phenomena like 2 above.