Proof that final topology with a certain property is unique

Question

Assume we are given a set of topological spaces $(X_i,\tau_i), \forall i \in I$, a set $Y$, a set of functions $f_i: X_i\rightarrow Y$, a topological space $(Z,\sigma)$ and a function $h : Y\rightarrow Z$.

Then assume that $h$ is continuous $\iff$ $h \circ f_i $ is continuous $\forall i \in I$.

Let $\tau$ be final topology on $Y$, defined $\tau = \{U \subset Y | f^{-1}_i (U) \in \tau_i, \forall i \in I\}$. I must prove that this topology is unique, ie. only topology on $Y$ that fulfills the requirement that $h$ is continuous $\iff$ $h \circ f_i $ is continuous $\forall i \in I$.

Attempt:

Assume that instead of $\tau$ we had $\tau^´$. Then assume that $g \in \sigma$. Now $(h \circ f_i)^{-1} (g) \in \tau_i,\ \forall i \in I$, for for continuous function, the preimage of an open set is open. Also $ f_i^{-1}(h^{-1}(g)) = f_i^{-1}(v), \ v \in \tau^´$, for the same reason.

Now $f_i^{-1}(v) \in \tau_i, \ \forall i \in I,$ for if they weren't, then $\tau_j \not\owns U=f_j^{-1}(v)=f_j^{-1}(h^{-1}(g))=(h \circ f_j)^{-1} (g) = U \in \tau_j$, for some $j \in I$, this is contradiction.

But what I cannot get out of my head are a few questions. Like, how can we know that there isn't some set $k \in \tau^´$ where $h (k) \notin \sigma$? This image $h (k)$ doesn't have to be closed, or does it? If it needs to be, then this case is violation of the continuity of $h$.

Also, how can we know that there is not some $t \subset Y$ in $\tau^´$ for which $f^{-1}_j(t) \notin \tau_j$ and it is not the preimage of any set in $\sigma$? This would be bigger than $\tau$ but we would have no way to get to these extra sets.

Answer 1

Like Arthur Fischer stated, you define $\tau$ as stated as a topology on $Y$ that depends on the topological spaces $(X_i, \tau_i)$ and $f_i$. Then the unicity statement should be: $\tau$ is the unique topology that has the property

$$\forall \mbox{ topological spaces } Z : \forall h: (Y,\tau) \rightarrow Z: ( h \mbox{ continuous } \iff \forall_{i \in I} \,(h \circ f_i) \mbox{ continuous. })$$

It is clear that $\tau$ satisfies this property, and you already know this judging from your question (it follows straight from the definition of $\tau$). Suppose that a topology $\tau'$ satisfies this property as well (we need to show that $\tau = \tau'$). Letting $h$ be the identity from $(Z,\tau')$ to $(Z, \tau)$ we see that by the fact that for all $i$ and all $O \in \tau$: $(h \circ f_i)^{-1}[O] = f^{-1}[O] \in \tau_i$, by the definition of $\tau$, so for all $i$, $h \circ f_i$ is continuous, and as $\tau'$ satisfies our desired property, $h$ is continuous, which means that $\tau \subset \tau'$, by the definition of continuity (of the identity map).

On the other hand, all $f_i$ are continuous as maps from $(X,\tau_i)$ to $(X,\tau)$, as this follows from the property as well, taking $h$ the identity on $(Y,\tau')$, which is always continuous (for any space), and $h \circ f_i = f_i$. But this means by definition that for any open set $O$ of $\tau'$ and any $i \in I$, $f^{-1}[O] \in \tau_i$, which just says that $O \in \tau$, and so $\tau' \subset \tau$, and we have equality and the unicity.

Often in this categorical proofs, going for "canonical" maps, like identities, is the key to success...