# Algebra objects of $\operatorname{Vec}(\mathscr{C})$ are lax functors $\mathscr{C}^\text{op}\to \operatorname{Vec}$

Let $$\mathscr{C}$$ be a rigid $$C^*$$-tensor category. Let $$\operatorname{Vec}(\mathscr{C})$$ be the category with linear functors $$\mathscr{C}^{\text{op}}\to \operatorname{Vec}$$ (= category of complex vector spaces) as objects and with natural transformations between these functors as morphisms. This is a monoidal category, see section 2.4 in the article "Operator algebras in rigid C*-tensor categories" by Jones and Penneys.

In this same article, the following is claimed:

I'm now trying to verify this claim. In neither direction, I see how this correspondence should work. So concretely, given an algebra object in $$\operatorname{Vec}(\mathscr{C})$$, how to associate a lax tensor functor $$\mathscr{C}^{\text{op}}\to \operatorname{Vec}$$ and conversely?

If $$\mathbf{A}$$ is an algebra object in $$\operatorname{Vec}(\mathcal{C})$$, then we have for every object $$c$$ maps $$\bigoplus_{a,b} \mathbf{A}(a)\otimes \mathcal{C}(c, a\otimes b)\otimes \mathbf{A}(b) \to \mathbf{A}(c)$$ where the direct sum ranges over simple objects. Taking $$x,y$$ simple, we consider the composition $$\mathbf{A}(x)\otimes \mathbf{A}(y)\hookrightarrow \bigoplus_{a,b} \mathbf{A}(a)\otimes \mathcal{C}(x\otimes y, a\otimes b)\otimes \mathbf{A}(b) \to \mathbf{A}(x\otimes y).$$ This gives us morphisms $$\mathbf{A}(x)\otimes \mathbf{A}(y)\to \mathbf{A}(x\otimes y)$$. There is a unique natural transformation between the bilinear functors $$\otimes \circ (\mathbf{A}\times \mathbf{A})\to \mathbf{A}\circ \otimes$$ with these values on the simple objects. This gives the desired Lax-tensor structure.
Conversely, if $$\mathbf{A}: \mathcal{C}^{\text{op}}\to \operatorname{Vec}$$ is a lax functor, with maps $$\mathbf{A}(x)\otimes \mathbf{A}(y)\to \mathbf{A}(x\otimes y)$$, consider the composition $$\bigoplus_{a,b}\mathbf{A}(a)\otimes \mathcal{C}(c, a\otimes b)\otimes \mathbf{A}(b) \cong \bigoplus_{a,b}\mathbf{A}(a\otimes b) \otimes \mathcal{C}(c,a\otimes b) \to \mathbf{A}(c)$$ which gives maps $$(\mathbf{A}\otimes \mathbf{A})(c)\to \mathbf{A}(c)$$ which induce the multiplication $$\mathbf{A}\otimes \mathbf{A} \implies\mathbf{A}$$.