This question is about (not necessarily symmetric) monoidal categories enriched over a symmetric monoidal category $\mathcal{V}$. Assume that $\mathcal{V}$ is closed. You may also assume that $\mathcal{V}$ is (co)complete if you wish. 


If $k$ is a commutative ring, a $k$ algebra can be defined in two ways. Either as a $k$-module $R$ together with morphisms $k\rightarrow R$ and $R\otimes\_{k}R\rightarrow R$ satisfying the well-known laws, or as a ring homomorphism to the center $k\rightarrow Z(R)$.


Let's see what happens in the categorical context.


The tensor product of $\mathcal{V}$-enriched categories can be straightforwardly defined, see [Kelly's book](http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html). Then one can define what a monoidal $\mathcal{V}$-category is by reproducing the classical definition in the enriched context.


Assume now that $\mathcal{C}$ is an ordinary monoidal category. I believe that the braided center $Z(\mathcal{C})$ as defined by [Joyal and Street](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V0K-45F5WD4-1N&_user=10&_coverDate=04%2F29%2F1991&_rdoc=1&_fmt=high&_orig=search&_origin=search&_sort=d&_docanchor=&view=c&_searchStrId=1603718721&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=efe9c8958bd0e8febd49ae40826f6f84&searchtype=a) is a well known construction. Suppose that we have a strong braided monoidal functor $z : \mathcal{V}\rightarrow Z(\mathcal{C})$ such that the functor $z(-)\otimes Y : \mathcal{V}\rightarrow \mathcal{C}$ has a right adjoint ${Hom}\_{\mathcal{C}}(Y,-) : \mathcal{C}\rightarrow\mathcal{V}$ for any object $Y$ in $\mathcal{C}$. The counit is an evaluation morphism in $\mathcal{C}$, 


$ev: z( {Hom}\_{\mathcal{C}}(Y,Z))\otimes Y\longrightarrow Z$


One can define composition morphisms in $\mathcal{V}$


${Hom}(Y,Z)\otimes {Hom}(X,Y)\longrightarrow {Hom}\_{\mathcal{C}}(X,Z) $


as the adjoint of


$z({Hom}(Y,Z)\otimes {Hom}(X,Y))\otimes X \cong
z({Hom}(Y,Z))\otimes z({Hom}(X,Y))\otimes X
\stackrel{id \otimes ev}\longrightarrow
z({Hom}(Y,Z))\otimes Y
\stackrel{ev}\longrightarrow
Z $


I think it's pretty obvious that $\mathcal{C}$ becomes $\mathcal{V}$-enriched in this way. Moreover, one can also enrich the tensor product in $\mathcal{C}$ in a similar way. 


Do you guys agree? Do you know of any reference where this is checked with some detail? Is it even more obvious than I think?


Any comment is welcome!



Enriched monoidal categories
There is a theorem in category theory, generally regarded as folklore, which says that for a symmetric monoidal closed category $V$, the following structures are equivalent:


1. a category $C$ with an action $V\times C\to C$ of the monoidal category $V$ on $C$, which we may write as $(v,c)\mapsto v\*c$, for which $-\*c:V\to C$ has a right adjoint for each $c\in C$ (here the action amounts to a strong monoidal functor $V\to [C,C]$.
2. a $V$-category $C$ for which the $V$-functor $C(c,-):C\to V$ has a left adjoint for each $c\in C$. (such a $V$-category is said to be "tensored'' or "copowered'')


You can see this, for example, in the appendix to [this paper](http://www.tac.mta.ca/tac/volumes/9/n4/n4.pdf).


In your case, unless I've misunderstood, the centre $Z(C)$ plays little role. The point is that your functor $z:V\to C$ induces an action via $v\*c=z(v)\otimes c$, and $-\*c$ has a right adjoint by assumption, so you get the $V$-enrichment.


(There is an analogous characterization of $V$-categories $C$ which are cotensored/powered: this means that each $C(-,d):C^{op}\to V$ has a left adjoint.)


