TITLE: Must the left and right unitors of a monoidal category coincide at the neutral object?
QUESTION [5 upvotes]: A monoidal category has (among other things) a pair of natural transformations called the left and right unitors
$$ \lambda_A:I\otimes A \cong A $$
  $$ \rho_A:A\otimes I \cong A $$
The categories I work with on a daily basis have $\lambda_I=\rho_I:I\otimes I \cong I$, but I think that is mostly a consequence of the fact that most of them have a terminal object as their $I$.  Is this true in general?
I'm sort of hoping this is true, because otherwise it seems that some of the definitions of enriched categories need to make an arbitrary choice of one unitor or the other.  For example, see display (1.10) of Max Kelly's introduction to enriched category theory on page 10.
(I'm actually interested in the even-more-general case of premonoidal categories, but mathematicians don't use those very often, so I've posed the question in terms of monoidal categories instead).

REPLY [6 votes]: In Categories for the Working Mathematician MacLane included $\lambda_I = \rho_I$ as one of three diagrams involving the associator and the two unitors that must commute (the other two being the usual pentagon and triangle diagrams) as the axioms defining monoidal categories.
It was however proven to follow from the other axioms in Max Kelly's 1964 paper On MacLane's Conditions for Coherence of Natural Associativities, Commutativities, etc. (Journal of Algebra 1, 397–402)
EDIT: For the case of premonoidal categories, I think $\lambda_I = \rho_I$ follows from the results of section 4 of John Power's Premonoidal categories as categories with algebraic structure (Theoretical Computer Science 278, 1-2, 303-321). In that paper, the unitors in a premonoidal category are defined to be central natural transformations; further down, the centre of a premonoidal category is defined to be the category with the same objects but with only the central morphisms of the original category, and this turns out to be a monoidal category, hence reducing to Kelly's proof.