About the universal property of initial and final topologies

As referenced by Theoretical Economist, an answer by Henno Brandsma indicates that the other inclusion can be proved by considering the identity as well, but changing the topologies in the domain/codomain.

In the case of $\tau$ being the initial topology, let $\tau'$ be a topology so that for any function $h$ we have that $h$ continuous if and only if $f_ih$ is continuous for all $i$. As I stated in the post it is easy to see that any other topology satisfying this is finer than the initial topology: the identity on $(X, \tau')$ makes $f_i : (X,\tau') \rightarrow Y_i$ continuous for all $i$ and so $\tau'$ contains the inital topology, generated by preimages of open sets of the $Y_i$ via $\{f_i\}_i$. To see that $\tau' \subseteq \tau$, it is equivalent to prove that $(X,\tau) \xrightarrow{id} (X,\tau')$ is continuous. Equivalently, we can see that each composition

$$ (X,\tau) \xrightarrow{id} (X,\tau') \xrightarrow{f_i} Y_i $$

is continuous, but these are the mappings $f_i : (X,\tau) \rightarrow Y_i$ which by definition of $\tau$ ought to be continuous, thus proving the original claim.

The case for the final topology follows likewise.

Let it be that $\tau$ is the topology on $X$ initialized by the $f_i$ in the sense that a function $h:Z\to X$ is continuous iff $f_ih:Z\to X_i$ for every $i\in I$.

Further let it be that $\tau'$ is the topology generated by the collection $\mathcal V=\{f_i^{-1}(U_i)\mid i\in I, U_i\in\tau_i\}$.

Then for every space $(Z,\tau_Z)$ and every function $h:Z\to X$ the following statements are equivalent:

• $h:(Z,\tau_Z)\to (X,\tau')$ is continuous
• $h^{-1}(f_i^{-1}(U_i))\subseteq\tau_Z$ for every $i\in I$ and every $U_i\in\tau_i$

• $f_ih:(Z,\tau_Z)\to (X,\tau_i)$ is continuous for every $i\in I$

• $h:(Z,\tau_Z)\to (X,\tau)$ is continuous.

Applying this on $\mathsf{id}:(X,\tau)\to(X,\tau')$ and $\mathsf{id}:(X,\tau')\to(X,\tau)$ we find that $\tau=\tau'$