Vector spaces are not rigid 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: 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.
