Maybe the title doesn't explain very well my question. I was wondering why if some object satisfies an universal property, then for any other object there is one and only one mapping between them. Let me explain myself a little more with an example. Suppose that we have three objects (I think we can suppose they are sets) $A,B$ and $C$ and two maps: $i:A\rightarrow B$ and $j:A\rightarrow C$. Let $h:B\rightarrow C$ be map such that makes commutative the diagram, i.e. $i\circ h = j $ (I use the opposite notation for composition). Under what assumptions for $A,B$ and $C$ and $i$ and $j$ such an $h$ is unique, that is, if $h'$ is another map satisfying $i\circ h' = j$ then $h=h'$.
As you can imagine, the background for this question is universal properties that arises in category theory: products, coproducts, pullbacks, pushouts, etc. can be defined by mean of them, and usually we find statements like for example the product: The product of two objects $A$ and $B$ is (whenever it exists) an object $C$ together with two maps $p:C\rightarrow A$ and $q:C\rightarrow B$ such that for any other object $D$ and arrows $f:D\rightarrow A$ and $g:D\rightarrow B$ theres is only one morphisms $h:D\rightarrow C$ such that $h\circ p = f$ and $h\circ q=g$.
I guess that the answer may be related with the span of $A$ and $B$ by $p$ and $q$ (or by the spant of $C$ by $i$ and $j$ if we have a corpoduct). Of course, when I say 'span' I'm not thinking in the usual meaning of the word span in linear algebra. I'm meaning... I don't know what exactly. Maube for Set a 'span' could be a surjective function, because then every object of $C$ would be written in terms of the elements of $A$ and $B$ (I'm thinking in coproducts here). Of course for categories like Ab span could mean span actually: in the case of direct sums, $i(a)$ and $j(b)$ generates the group $A\oplus B$.
I hope the answer would be clear and interesting and you understand me.
Thanks in advance
EDIT
To sum up: I'm not asking why $h$ must be unqiue, but why it is. Which is the reason(s) behind $A,B,C,p$ and $q$ (or $i$ and $j$) for make $h$ unique.
For the product (coproduct) of two objects, it can be define as a final object in the comma category. Then, by definition of final (initial) object, $h$ is unique. I suppose it can be extend to an arbitrary family of objects, but I haven't got enough knowledge. But that would be another answer. Then, my questions would be, what about pullbacks and pushouts and direct and inverse limits?
EDIT 2: (Explanation for 'span')
I'm studying at this moment the pushout of two maps in Set and I think it is a good example to explain what I mean by ''span''. as you know perfectly, the pushout of two maps $f:X\rightarrow A$ and $g:X\rightarrow B$ is the set $A\coprod B$ quotient by the finest equivalence relation $(f(x),0)\sim (g(x),1)$ and the maps $i_A=\iota_A\circ \pi$ and $i_B=\iota_B \circ \pi$, where $\pi$ is the canonical projection onto $A\coprod B/\sim$ and $\iota_A$ and $\iota_B$ are the inclusions of $A$ and $B$ into the disjoint union. Let me call $Y=A\coprod B/\sim$.
For any other set $Z$ and maps $j_A:A\rightarrow Z$ and $j_B:B\rightarrow Z$ satisfying $f\circ j_A=g\circ j_B$ I define a map $h:Y\rightarrow Z$. You know better than me how this map must be defined. So come on to the interesting part: this pam $h$ satisfies $i_A\circ h = j_A$ and $i_B\circ h = j_B$. And now, if $h':Y\rightarrow Z$ is any other map satisfying the same conditions, then $h=h'$, and that is because for any $y\in Y$, $y$ must be of the form $i_A(a)$ or $i_B(b)$ (maybe both at the same time) so the commuting condition defines $h'$ for all $y\in Y$.
A similar case happens in the corpoduct of two abelian groups for example. Every element of $A\oplus B$ can be express as the product of $\iota_A(a)$ and $\iota_B(b)$, so again the commuting relations determine the map completely.
A way to define what is a universal mapping problem is through representability.
A functor $F : \mathcal C^{\rm op}\to \mathbf{Set}$ is said to be representable if there exists $X\in\mathcal C$ and a natural isomorphism $$ \alpha : F \to \mathcal C(-,X) $$ Denote $x$ the unique element of $F(X)$ such that $\alpha_X(x) = \mathrm{id}_X$. Now the naturality of $\alpha$ gives the following property:
Proof. By naturality of $\alpha$, and for any map $k: Z \to X$, we have $k = \alpha_Z(F(k)(x))$. Denote $h = \alpha_Y(y)$, then it gives $$ \alpha_Y(y) = \alpha_Y(F(h)(x)) $$ But $\alpha_Y$ is injective, so $y = F(h)(x)$ and $h$ answer to the problem. Now if there is another $h':Y \to X$ such that $F(h')(x)= y$, then $h' = \alpha_Y(F(h')(x)) = \alpha_Y(y) = h$. This proves the uniqueness of $h$.
So the answer to "why is $h$ unique" is that universal mapping problem requires an $\alpha$ which is both natural and bijective in each component. Remove either of this condition to $\alpha$ and you would have a "generalized notion of universal mapping problem" where the uniqueness of $h$ is no longer required. (Also, this generalized version would be clearly less interesting, but that is another issue.)
Here is the link of the above presentation with the examples you gave in you post.
Given a category $\mathcal A$, if you take $\mathcal C = \mathcal A$ and $F = \mathcal C(-,A) \times \mathcal C(-,B)$ for some fixed $A,B$, then an $X$ representing $F$ is actually a product for $A,B$ and the $x\in F(X)$ is given by the projections $(\pi_A,\pi_B)$.
Given a category $\mathcal A$, if you take $\mathcal C = \mathcal A^{\rm op}$ and $F = \mathcal A(A,-) \times \mathcal (B,-)$ for some fixed $A,B)$, then an $X$ representing $F$ is a coproduct for $A,B$ and the $x\in F(X)$ is given by the inclusions $(i_A,i_B)$.