# Uniqueness of the initial topology

The Initial Topology $\tau_X$(as defined below) and its uniqueness is characterised by two certain properties.

I would like to know, what happens to the uniqueness of $\tau_X$, if you remove one or even both properties? Possibly $\tau_X$ is then no longer unique. How would/could such topologies then look like?

Given: A set $X$ and $(Y_i)_{i \in \ I}$ topological spaces and maps $f_i : X \rightarrow Y_i$

There is then an unique topology $\tau_X$ with:

1. All $f_i$ are continuous
2. Given a topological space $Z$ and a map $g: Z \rightarrow X$, then: $g$ continuous $\Leftrightarrow f_i \circ g: Z \rightarrow Y_i$ continuous $\forall i \in I$

$\tau_X$ defined above is called the Initial Topology or the Cofinal Topology with respect to the $f_i$.

• Well, if you do not assume 2, then any topology on $X$ finer than $\tau_X$ is possible. – cjackal Jul 7 '18 at 10:26

If you just assume 1. we could just take the discrete topology on $X$ and this would make all functions continuous on it.
Usually the initial topology is defined as the intersection of all topologies that make all $f_i$ continuous. This intersection is non-empty (the discrete topology is one of these topologies as we saw) and the intersection of topologies is again a topology. The property 2. then follows, once you observe that we can also say that $X$ is the topology generated by the subbase $\mathcal{S}=\{f_i^{-1}[O]: i \in I, O \subseteq Y_i\}$.
The nice part is that property 2 implies that $X$ must carry this topology. I write more on this in this answer.
In fact, property 2. implies property 1, because $g= \mathrm{id}_X$ is always continuous. So just property 2 alone characterises the initial topology.