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?

  • $\begingroup$ @celtschk I totally had a typo there, I've fixed it. $\endgroup$ – Guido A. Sep 15 '18 at 9:28
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    $\begingroup$ Ok, now it makes sense. $\endgroup$ – celtschk Sep 15 '18 at 9:30
  • $\begingroup$ "which is the coarser so that the functions" -- this seems like a typo? $\endgroup$ – Theoretical Economist Sep 15 '18 at 9:46
  • $\begingroup$ @TheoreticalEconomist gramatically or conceptually? I tried to say that this is the topology with 'less open sets' so that every function in the family is continuous. $\endgroup$ – Guido A. Sep 15 '18 at 9:47
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    $\begingroup$ Grammatically. I think you mean “which is the coarsest topology such that the functions”? $\endgroup$ – Theoretical Economist Sep 15 '18 at 9:49

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 $\tau=\tau'$


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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