What is the intuition behind the uniqueness constraint for morphisms determining products? I am following Bartosz Milewski's book and weblectures and have a question about the definition of the categorical product.
He uses names that are a bit different from the ones I can find on Google, so here is the definition. An object $c$ is the product of objects $a$ and $b$ (just denoted $c$, not $a \times b$) if it has the projections $p : c \rightarrow a$ and $q : c \rightarrow b$ such that for any object $c'$ together with projections $p' : c' \rightarrow a$ and $q' : c' \rightarrow b$ there exists an unique morphism $m : c' \rightarrow c$ that make the triangle commute, i.e. such that $p \cdot m = p'$ and $q \cdot m = q'$, i.e. "there exists only one of them that makes the triangles commute". 
Now, what is the intuition behind this uniqueness constraint? I'm not looking for a counterexample, but for the "reason" behind it. What happens when we drop the uniqueness constraint? I know that if we do we satisfy all requirements for $a = a, b = b, c = a \times a \times b, c' = a \times b$, which we should not be able to. When we drop the uniqueness constraint, we allow for the situation that the morphism $m$ "makes up" something by selecting some random value for the unused $a$.
Can we state something about this in terms of bounds or entropy? I.e. something like 'a product is the "lowest-information situation" possible while still having all information available to be general enough to always satisfy the laws' or something like that? Like some greatest lower bound? An by enforcing $m$ to be unique we ensure that it cannot "add entropy/information" in any way?
Edit: I guess what I'm asking is: what is the consequence of one object having an unique mapping to another? If a morphism $a \rightarrow b$ is unique, what does this tell us about the information content of $a$ and $b$? Surely $a$ is "bigger"/contains more information than $b$, right? Doesn't $b$ even have to be a singleton in this sitution?
Edit2: OK let me reformulate my question. When we look at the object that forms the product, we are looking at the /smallest possible object/ that still satisfies all laws. Any bigger object we can "shave down" to this most basic object, and still retrieve the two components. Dually, when defining a sum, we want the most general object possible, any instantiation we can generalize. We define these greatest lower bounds and least upper bounds trough saying there must be an unique morphism between them. My question: how does the uniqueness of a morphism say anything about the size of the objects at its beginning and end?
 A: When you define the product (not a product) of $X$ and $Y$, you are in fact defining 4 things at the same time:


*

*$X \times Y$

*$\pi_1 : X \times Y \; \to \; X$

*$\pi_2 : X \times Y \; \to \; Y$

*$\langle f, g \rangle$, for any $Z$, $f : Z \to X$ and $g : Z \to Y$
Those things must be unique or else this wouldn't be a definition for the product.
The unicity for the second, third and fourth things is expressed with equality. While the unicity for the first thing is expressed with isomorphism, because in category theory we are only interested in objects up to isomorphism.
A: The idea is that a map into a product should be entirely determined by its coordinates. 
Why do we want that ? Well products are supposed to generalize the usual cartesian products of sets, groups, topological products, where an element is indeed entirely determined by its coordinates; and therefore so are the maps into the product. 
You can check out my answer here, which is more generally about why we define the product the way we do, in particular this should motivate the uniqueness. 
But I think the "points in a product have coordinates that determine them"-view is a good way to see categorical products
A: I want to address your edits first. A priori, categories have no notion of size, entropy, or information. You would need to formulate such a thing.
One possibility is by considering presheaves and the Yoneda embedding. I realize that these concepts will likely be unfamiliar to you right now, but they may help future visitors, including potentially yourself.
First reason for why we define products this way
Because this is the property that the Cartesian product of sets satisfies.
Brief review of presheaves and the Yoneda embedding
A presheaf on a category $\newcommand\C{\mathcal{C}}\newcommand\op{\text{op}}\C$ is a functor $\C^\op\to \newcommand\Set{\mathbf{Set}}\Set$. 
The category of presheaves on $\C$ has presheaves for objects and natural transformations as morphisms, and is denoted $[\C^\op,\Set]$. 
There is a canonical fully-faithful functor $y:\C\to [\C^\op,\Set]$ defined by 
$y(x) = \C(-,x)$.
We can therefore view $y(x)$ as translating an abstract object into a collection of sets that we can measure the size of, and thus regard this as a sort of measure of the information contained in $x$.
Back to products
If $c$ has morphisms $p:c\to a$, and $q:c\to b$, then we can define a morphism
$y(c)\to y(a)\times y(b)$, where $y(a)\times y(b)$ means the pointwise cartesian product of sets (which is also the categorical product of the functors). 
The morphism is defined as follows. Given $f\in y(c)(x) = \C(x,c)$, 
$p\circ f : x\to a$ and $q\circ f : x\to b$, so defining 
$f\mapsto (pf,qf)$ makes sense, and you can check that this is in fact natural.
Call this morphism $\phi_{p,q} : y(c)\to y(a)\times y(b)$. 
The existence requirement translates into the statement that $\phi_{p,q}$ is (pointwise) surjective, and the uniqueness requirement translates into the statement that $\phi_{p,q}$ is (pointwise) injective. The two together become the statement that $\phi_{p,q}$ is an isomorphism. In a metaphorical sense, $c$ encodes the cartesian product of the information of $a$ and $b$.
A: Have a look at Chapter 7 of https://www.logicmatters.net/resources/pdfs/GentleIntro.pdf . In §7.2 I spend some time pre-categorially motivating a general idea of what makes for (something that will play the role of) a product, which in turn motivates §7.3 which gives the standard categorial definition of a product in a very natural way, including that uniqueness requirement.
A: I'm aware you're not looking for a counterexample, but the following is too long for a comment and might still be an interesting case to consider in the context of this question:
Consider the category of probabilistic mappings, where objects are measure spaces $X,Y,Z,\ldots$ and morphisms are Markov kernels / probabilistic mappings: So for $f: X\to Y$, $f(x)$ isn't necessarily uniquely defined, but rather a probability distribution on $Y$.
First observation: What are points in this category? A morphism $f: \ast\to X$ is simply a probability distribution on $X$.
Second observation: The category encompasses the category of measure spaces since any deterministic mapping can be viewed as a probabilistic mapping.
Now, consider the product measure space $X\times Y$ with its projections onto $X,Y$. Does this turn it into a product in the category of probabilistic mappings? For this to be true, we should have $\text{Hom}(\ast,X\times Y)\stackrel{\cong}{\to}\text{Hom}(\ast,X)\times\text{Hom}(\ast,Y)$, which would mean that probability distributions on $X\times Y$ are in 1-1 correspondence with pairs of probability distributions on $X,Y$. 
That's of course not true: There are distributions on $X\times Y$ whose coordinate distributions aren't independent, and those don't come from a pair of probability distributions on $X,Y$. Yet, every pair of probability distributions on $X,Y$ comes from a probability distributions on $X\times Y$, namely the product distribution.
So $\text{Hom}(\ast,X\times Y){\to}\text{Hom}(\ast,X)\times\text{Hom}(\ast,Y)$ is indeed surjective, but not injective / uniqueness fails.
A: Dropping the uniqueness requirement means you won't be able to proof the product's uniqueness up to isomorphism. Indeed, let me sketch the proof that the product is, in fact, isomorphic to any other product.
Let $A$ and $B$ be some types. Assume $X, \pi^X_A, \pi^X_B$ and $Y, \pi^Y_A, \pi^Y_B$ are two products, i.e. $\pi^X_A \in Hom(X, A)$ and similarly for $\pi^X_B, \pi^Y_A, \pi^Y_B$. Additionally, call $m^X_M \in Hom(M, X)$ / $m^Y_M \in Hom(M, Y)$ the unique morphism into $X$/$Y$, for a type $M$ with projections into $A$ and $B$.
How to show that $X$ and $Y$ are isomorphic? Well, certainly, we get a morphism $X \to Y$ by the universal property for $Y$, i.e. $m^Y_X \in Hom(X, Y)$ and also $m^Y_X \in Hom(Y, X)$. What remains to be shown is that these two are mutual inverses: $m^Y_X \circ m^X_Y = id_Y$ and $m^X_Y \circ m^Y_X = id_X$.
Here is where uniqueness comes into play. Since both $m^X_Y \circ m^Y_X$ and $id_X$ are morphisms in $Hom(X, X)$ and $X$ itself fulfills the conditions to invoke the universal property and this property says that the morphism is unique we have, by the uniqueness of $m^X_X$:
$id_X = m^X_X = m^X_Y \circ m^Y_X$
Similarly we can proof $id_Y = m^Y_X \circ m^X_Y$, which concludes the proof that the two objects are isomorphic.
So far, only the uniqueness of $m^X_X$ and $m^Y_Y$ has been used. But the two products are not proven fully isomorphic yet.  What remains to be shown is that the structure of them, i.e. the projections and their universal property, agree when transporting along the isomorphism. For that we have to show:
$
\pi^X_A = \pi^Y_A \circ m^Y_X \\
\pi^X_B = \pi^Y_B \circ m^Y_X \\
m^Y_M = m^Y_X \circ m^X_M, \text{for all $M$ with correct shape}
$
The first two are easy, they are in fact given again by the universal property for $Y$, invoked for taking $M = X$. The last line, then, would be impossible, were it not for the uniqueness property that immediately let's us conclude the proof since $m^Y_X \circ m^X_M \in Hom(M, Y)$.
A: I thought about it some more, and have been able to come up with an explination such that it makes sense for me.
When we relate morphisms to set theory, if we have an unique $f :: a \to b$, it means $f$ is surjective: it uses the entire codomain. As functions cannot map one element in the domain to multiple elements in the codomain, it means that $a$ is at least as big as $b$. Now, if we require no further conditions, we eventually morph into the smallest set satisfying no conditions, i.e. the empty set. However, if we set some requirements, we eventually morph into the smallest set still satisfying the conditions. This is how a product is defined: the smallest set such that you can construct both parts again, i.e. their union without any extra information. Even more specific: all elements of $a$ and $b$ must be in $a \times b$ as otherwise the conditions will not hold. Furthermore, no additional elements can be inside $a \times b$ because otherwise there would be an unique mapping from it to the set that does not contain this additional element, i.e. it would not be the smallest.
The dual applies for the sum: here we are looking for the biggest set such that it still satisfies the conditions.
