If $Cob_n$ is $n$-dimensional cobordisms, and $Z \colon Cob_n \to \mathcal{V}$ is an $n$-dimensional TQFT (i.e. a symmetric monoidal functor - in particular, $\mathcal{V}$ is symmetric), then one can in principle define an $(n-1)$-dimensional TQFT $Z^\flat \colon Cob_{n-1} \to \mathcal{V}$. This process is called "dimensional reduction", and it works as follows: $Z^\flat(M) = Z(M \times S^1)$.
It feels like this defines a functor $-^\flat \colon$ {$n$-dimensional $\mathcal{V}$-valued TQFTS} $\to$ {$(n-1)$-dimensional $\mathcal{V}$-valued TQFTS}.
Question: Is this correct, and (when) does this functor have an adjoint?
This probably won't work as well if $Cob_n$ has more structure, but I'm just interested in this simplest case for now.