Introduction (a bit long)
I am very new to category theory and algebra, in general.
I recently saw the formalized concept of a universal morphism, here in WikipediA.
I will repeat the definition with my own words here for convenience.
Definition.
Let $\mathcal{C}$ and $\mathcal{D}$ be categories, let $F:\mathcal{C}\to\mathcal{D}$ be a functor and let $A \in \mathcal{C}, X \in \mathcal{D}$ be objects. We say that a triple $(u,A,X)$, where $u: X \to F(A)$, is a universal morphism if, and only if, the following property holds:
for any object $B \in \mathcal{C}$ and any morphism $f:X \to F(B)$, there is exactly one morphism $h_{f}:A \to B$ such that $f=F(h_{f})\circ u$, making the following diagram commute:
Okay. Because I wanted to understand it, I tried seeing how it applies to linear algebra: I heard that the following is a universal property:
If $V$ is a $\mathbb{K}$-vector space and $\mathcal{B}$ is a basis of $V$, then, for any $\mathbb{K}$-vector space $W$ and any function $f:\mathcal{B}\to W$, there is a unique linear map $\bar{f}:V\rightarrow W$ that extends $f$, that is, $\bar{f}_{∣\mathcal{B}}=f$.
In other words, the function $𝜓:\hom_{\text{Vect}_{\mathbb{K}}}(V,W)\rightarrow \hom_{\text{Set}}(B,W)$ such that $𝜓(g)=g_{∣\mathcal{B}}$ is the 'restriction to $\mathcal{B}$ function', is a bijection.
So this is actually a universal property because:
$\mathcal{C} =\text{Vect}_{\mathbb{K}}$ is the category of $\mathbb{K}$-vector spaces and $\mathcal{D}=\text{Set}$ is the category of sets and functions. The functor $F: \text{Vect}_{\mathbb{K}} \to \text{Set}$ is the forgetful functor. The object $X=\mathcal{B}$ is any basis of $V$ and $A=V$.
The universal morphism from $B$ to $F$ is just the set inclusion $B↪V$, so that we have that the following diagram commutes:
Here, I have made an abuse of notation and I have not written the forgetful functor except on top of the drawing.
Returning to the main point
Now, my goal is to also acknowledge the axiomatic definition of the tensor product as a particular case of the formalization of the universal property described before. Like, justify the reason for saying that the definition uses a universal property.
Definition.
Let $E_{1},…,E_{n}$ be $\mathbb{K}$-vector spaces. We say that a pair $(T,\varphi)$, consisting of a $\mathbb{K}$-vector space $T$ and an $n$-multilinear function $\varphi$ is a tensor product of the spaces $E_{1},…,E_{n}$ if, and only if, the following property holds:
for any $\mathbb{K}$-vector space $W$ and any $n$-multilinear map $f:E_{1} \times … \times E_{n} \to W$, there is exactly one linear map $f:T \to W$ such that $f= \bar{f} \circ \varphi$, that is, the following diagram is commutative:
Question
What category should $\mathcal{D}$ be? I can clearly see that $\varphi$ should be the universal morphism $u$, $h_{f}=\bar{f}$, $X=E_{1} \times … \times E_{n}$, $A=T$, $\mathcal{C}$ should be the cateogory of $\mathbb{K}$-vector spaces and linear maps, but I am unsure about what functor is $F$ because I do not know what suitable category should $\mathcal{D}$ be.
I observe that in $\mathcal{D}$, we have that a morphism should be an $n$-multilinear map, because $f$ and $𝜑$ are, but then, for the diagram to commute, the arrow $ \bar{f} : T \to W$ is linear, so it does not seem like it should be a morphism in $\mathcal{D}$.
Also, it seems like $\mathcal{D}$ could be the category of sets and functions, as the example from before in the linear case, but then it does not work because we are only considering $n$multilinear functions $E_{1}×…×E_{n} \to W$, not any ordinary function.
Best Answer
It's a good question. I would personally say that this definition of universal property is not the best one and is not what comes naturally in many situations.
On the other hand, I can tell you one easy way to formalize the universal property of tensor products: given $E_1,\dots, E_n$, there is a functor $M:\mathrm{Vect}_K\to \mathrm{Sets}$ which sends a space $V$ to the set of $n$-linear maps $E_1\times\cdots\times E_n\to V$ (the functor structure is clear: composing a multilinear map with a linear one gives a multilinear map, so the action on morphisms is just the obvious push-forward). Then the tensor product $E_1\otimes\cdots\otimes E_n$ represents this functor $M$.
The definition you quote basically says that $A$ should represent a functor of the form $\operatorname{Hom}_\mathcal{D}(X,F(\bullet))$ for some $X\in \mathcal{D}$ and some $F:\mathcal{C}\to \mathcal{D}$. It's actually always possible to reduce to this case, because any functor $M:\mathcal{C}\to \mathrm{Sets}$ can be represented in that form: worst case scenario, take $\mathcal{D}=\mathrm{Sets}$, $X=\{\ast\}$ (any singleton), and $F=M$; then $\operatorname{Hom}_\mathcal{D}(X,F(\bullet))$ is naturally isomorphic to $M$. But obviously this is a bit cumbersome.
So in my opinion, while the definition on wikipedia is not wrong, in the example you bring forth it is not the most natural one to use, and it is more natural to just refer to the general notion of representing objects for functors. It's not strictly speaking more general, but it's certainly more flexible.