Im following Etingofs Tensor categories and have read about rigid categories now.
There it says
The category of all vector spaces (including infinite dimensional) is not rigid. Take $V$ to be infinite dimensional. "Suppose that $c: k \rightarrow V \otimes Y$ is a coevaluation and take the subspace $V'$ of $V$ spannend by the first component of $c(1)$. This subspace is finite dimensional and yet the composition $V \rightarrow V \otimes Y \otimes V \rightarrow V$ which is supposed to be the identity map, lands in $V'$ - a contradiction."
It may be super easy, but I don't quite get it. What kind of composition is he talking about? (Which maps can be written over the arrows?) Why does this land in $V'$?
Thanks in advance!
A rigid category is supposed assign to each object $V$ a dual object $Y$ and morphisms $c: 1\rightarrow V\otimes Y$ and $d:Y\otimes V \rightarrow 1$ such that a certain composition of maps yields the identity. (I'm using $1$ here generically to denote the monoidal unit; in the case of vector spaces, the monoidal unit is the field $k$ .)
Specifically, both of the following transformations are supposed to yield the identity:
$$V \xrightarrow{c \otimes \text{id}} V\otimes Y \otimes V \xrightarrow{\text{id}\otimes d} V$$
and
$$Y \xrightarrow{\text{id} \otimes c} V\otimes Y \otimes V \xrightarrow{d\otimes\text{id}} V$$
Unfortunately, this is possible only if we restrict our attention to finite dimensional vector spaces, not infinite-dimensional vector spaces: suppose $X$ is an infinite dimensional vector space with dual $Y$ and maps $c,d$. We can prove that the above transformations are not the identity.
The function $c$ is a linear map from the underlying field $k$ into $V\otimes Y$. It is linear, so watch where $c(1)$ goes: it goes into the tensor product of a subspace in $V$ and a subspace in $Y$. Both of these are finite dimensional subspaces because $k$ is finite-dimensional.
So we have $V \xrightarrow{c\otimes\text{id}} V^\prime \otimes Y^\prime \otimes V \rightarrow V^\prime $
where $V^\prime$ and $Y^\prime$ are finite-dimensional. Evidently this map is not the identity because $V$ isn't finite dimensional.