# does an ordinary map have a universal property?

I use Lang's Algebra book to define universal objects:

Let $$\mathcal{C}$$ be a category. An object $$P$$ of $$\mathcal{C}$$ is called universal attracting if there exists a unique morphism of each object of $$\mathcal{C}$$ into $$P$$, and is called universal repelling if for every object of $$\mathcal{C}$$ there exists a unique morphism of $$P$$ into this object.

This definition is simple and easy to understand. I can analyze things like a quotient group and understand its universal property. For example, Let $$H$$ be a normal subgroup of $$G$$, any homomorphism $$f:G\to K$$ whose kernel contains $$H$$ factors over $$\pi: G\to G/H$$. That is, there is a unique $$g: G/H \to K$$ such that $$f=g\circ \pi$$. The category of this scene has objects of all homomorphisms $$f: G\to K$$ whose kernel contains $$H$$, and has morphisms $$g: K_1\mapsto K_2$$ such that $$g\circ f_1=g(f_1)=f_2$$. And $$\pi$$ is the universal repelling object of this category, since for any object $$f$$, there is a unique morphism $$g$$ sending $$\pi$$ to $$f$$: $$g\circ\pi=f$$.

I noticed that universal properties were usually organized by wording like "For each $$x\in X$$, there is a unique $$y\in Y$$, such that some properties are held for $$x$$ and $$y$$". And I also noticed for an ordinary map $$f:X\to Y$$, we had a similar statement:

For each $$x\in X$$, there exists a unique $$y\in Y$$, such that $$f(x)=y$$.

It looks like there is some category and a universal object. So I tried to construct but failed. My attempts include treating the map as a bipartie graph with arrows from $$x$$ to $$f(x)$$, treating these vertexes as objects and arrows as morphisms. I added an extra vertex $$e$$, who has all other vertexes pointing to it. Then $$e$$ is obviously a universal attracting object, but I don't know how to relate the above "For each... there exists a unique..." statement with this construction.

Any hints are appreciated.

I think that you're confusing the objects of a category with sets, and morphisms with functions. It doesn't make sense to say "For each $$x \in X$$", for an object $$X$$ of a category $$\mathscr C$$, because objects do not have elements in general. Likewise, morphisms are not functions and are not applied to elements. So the statement $$f(x) = y$$ is not well-formed.
That said, I will mention a result that seems somewhat related. The condition you describe looks similar to the property of a binary relation being functional. It turns out that such relations do have a universal characterisation. In particular, in the bicategory of sets and relations, $$\mathbf{Rel}$$, the relations that are functional are precisely those that are left adjoints, which are intimately related to universal properties. (I'm glossing over some details here: really this is a more general notion of "adjoint" than an adjoint functor, but hopefully it gives you an idea of the connection.)