General Topology – How to Understand the Universal Property of Initial and Final Topologies

general-topology

I have recently seen in my topology course that if $X$ is any set,

  1. Given a family of functions $\{f_i : X \to Y_i\}_{i \in I}$, with $(Y_i, \tau_i)_i$ topological spaces, the initial topology on $X$ with respect to this family is the one generated by $\{f_i^{-1}(U_i) : i \in I, U_i \in \tau_i \}$, which is the coarsest such that the functions $f_i$ are continuous, and verifies $h : Z \to X$ continuous if and only if $f_ih$ is continuous for all $i$ in $I$.

  2. In a similar way, given a family $\{f_i : Y_i \to X\}_{i \in I}$, the final topology $\tau$ in $X$ is defined by $U \in \tau $ if and only if $f_i^{-1}(U) \in \tau_i \ (\forall i \in I)$. This is the finest topology such that the family is countinuous, and $h : X \to Z$ is continuous if and only if $hf_i$ is for all $i \in I$.

It was also mentioned that the reciprocal of both statements is true, i.e. that if a topological space $(X, \tau)$ verifies $h : Z \to X$ (resp. $h:X \to Z$) continuous if and only if $f_ih$ (resp. $hf_i$) continuous for all $i$, for all $h$, it has the initial (resp. final) topology with respect to the family $(f_i)_i$.

It is easy to see taking $h \equiv id $ that the given topology $\tau$ is finer/coarser than the initial/final topology, because all the functions $f_i$ are continuous.

Any hints on how to prove the other inclusion?

Best Answer

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.

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