Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Is the following what you want to show: $\tau$ is the unique topology on $Y$ such that for any topological space $Z$ a mapping $h : Y \to Z$ is continuous iff $f_i \circ h : X_i \to Z$ is continuous for all $i \in I$. –  Arthur Fischer Jan 28 '13 at 17:42
    
@ArthurFischer Z is given and its $h \circ f_i$. Otherwise I think you got it. –  Valtteri Jan 28 '13 at 17:48
1  
Oops... Stupid mistake on the order of the composition on my part. But I don't think that what you want to show is true. If, for example, $\sigma$ was the trivial (anti-discrete) topology on $Z$ then every mapping into $Z$ is continuous, and thus the "iff" holds for every topology on $Y$. –  Arthur Fischer Jan 28 '13 at 18:08
    
@ArthurFischer Hmmmm, all the material comes from this book: math.ru.nl/~mueger/topology2012.pdf , page 35, definiton 5.3.6 , exercise 5.3.7 and proposition 5.3.8. The task is to show that the topology $\tau$ mentioned in the first sentence of 5.3.8 is only one that fulfills the requirement in the second sentence of 5.3.8. –  Valtteri Jan 28 '13 at 18:17
1  
@Arthur: It’s clearly supposed to be a statement of a universal property, so it should be for all $Z$, as in your original comment. –  Brian M. Scott Jan 28 '13 at 19:59
show 3 more comments

1 Answer

up vote 2 down vote accepted

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...

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.