In class, we had to state the universal property of $\text{Bilin}(A,A',-):\text{Ab}\to\text{Set}$, the functor taking an abelian group $B$ to the set of bilinear maps $f:U(A)\times U(A')\to U(B)$ where $U:\text{Ab}\to\text{Set}$ is the forgetful functor. This was the solution given (under the assumption that some representing object $A\otimes A'$ of $\text{Bilin}(A,A',-)$ exists):
Let $\alpha:H^{A\otimes A'}\cong\text{Bilin}(A,A',-)$ be the isomorphism relating the representing object to the functor. Define $\tau:=\alpha(A\otimes A')(\text{id}_{A\otimes A'})$, then for all $X\in\text{Ab}_0$ and $g\in\text{Bilin}(A,A',X)$ there exists a unique $\bar{g}\in\text{Ab}(A\otimes A',X)$ such that $g=(\text{Bilin}(A,A',-)(\bar{g}))(\tau)$.
My question: how exactly is this a universal property? nLab states that a universal property boils down to the object (or something closely associated) is initial in an appropriate auxiliary category.
What I could imagine is that this means we look at the category of dependent pairs $(X,g)\in\sum_{\text{Ab}_0}\text{Bilin}(A,A',X)$ with morphisms $\bar{f}:(X,g)\to (Y,f\circ g)$ for any $f:X\to Y$ (is there some fancy description of this construct, maybe in terms of comma categories?) and then $(A\otimes A',\tau)$ is initial.
But what interests me now, and maybe this is completely obvious, but is $\text{Bilin}(A,A',-)$ also initial in it's own isomorphism class in $[\text{Ab},\text{Set}]$?
I'm sorry if this all is just confused gibberish. Sometimes category theory blurs my thinking way too much.