I was scrolling nlab instead of studying my topology final, and stumbled on the following page: https://ncatlab.org/nlab/show/separation+axioms+in+terms+of+lifting+properties#hausdorff_spaces_
And intrigued, I tried to understand the lifting property for Hausdorff spaces. Which I understand as follows:
Denote by $*$ the singleton topological space, by 2 the indiscrete space on two points $\{0,1\}$ and by A the space whose set is $\{0,1,2\}$ and whose open sets are $\{\emptyset, \{0\}, \{1\}, \{0,1\}, \{0,1,2\}\}$. We have the following categorical definition of X being Hausdorff: X is hausdorff is and only if for all $f:2\to X$, which makes the square commute, there exists a diagonal morphism, such that the diagram commutes 
I am almost sure I misunderstood this, because I am assuming all arrows should be continuous, but taking the indiscrete on 2, means the function from 2 to A just isn't. Should I just take the discrete instead of the indiscrete on 2? but that isn't what the nlab is leading me to believe. Am I misunderstanding the category in which this diagram lies?
Best Answer
If you change topology on $2$ to be discrete, and assume $f$ is restricted to be monomorphism, then this property is indeed saying that $X$ is Hausdorff. We are in the category of topological spaces, and all arrows are continuous functions. Notice also that $*$ is redundant here, as compositions of any functions that end up in $*$ are constant.
$f:2\to X$ means pick any 2 points in $X$. If exists $g:X\to A$ such that the diagram commutes, then $g(f(0)) = 0$ and $g(f(1)) = 1$. Therefore $g^{-1}(\{0\})$ and $g^{-1}(\{1\})$ are disjoint open neighborhoods of $f(0)$ and $f(1)$ respectively. So any two different points (monomorphism assumption) have disjoint neighborhoods.
Adding another element $2 \in A$ ensures that all Hausdorff spaces have this property. If we have $x, y \in X, x\neq y$ and their open disjoint neighborhoods $G_x, G_y$ then we can define $f(0) = x, f(1) = y$ and $$ g(a) = \begin{cases} 0 \quad : a \in G_x \\ 1 \quad : a \in G_y \\ 2 \quad : a \in (G_x\cup G_y)^c. \end{cases} $$ you can easily check that this function is continuous and diagram commutes.