Suppose $X$ is a topological space whose topology is coherent with a family $\mathcal{B}$ of subspaces. The universal property of coherence is this:

If $Y$ is another topological space, then a map $f : X \to Y$ is continuous if and only if $f|_{B}$ is continuous for every $B \in \mathcal{B}$.

I'm attempting to prove this

My Attempted Proof:

Suppose $f : X \to Y$ is continuous then $f|_{B}$ is trivially continuous by restricting the domain of $X$ to $B$.

Now suppose $f|_{B}$ is continuous for every $B \in \mathcal{B}$. Choose $B \in \mathcal{B}$ and observe that for every open $V$ in $Y$ we have $f|_{B}^{-1}[V]$ to be open in $B$ and since $B$ is a subspace of $X$ we thus have $f|_{B}^{-1}[V] = U_B \cap B$ for some open set $U_B$ in $X$.

Now pick an open $V$ in $Y$ and note that $\bigcup_{B \in \mathcal{B}} B = X$ and that $f|_{B}^{-1}[V] = \{x \in B \ | \ f(x) \in V\}$ then $$\bigcup_{B \in \mathcal{B}} f|_{B}^{-1}[V] = \bigcup_{B \in \mathcal{B}} \left(U_B \cap B\right) = U \cap X = U = \{x \in \bigcup_{B \in \mathcal{B}} B = X \ | \ f(x) \in V\} = f^{-1}[V]$$ where $U = \bigcup_{B \in \mathcal{B}} U_B$. Hence $f^{-1}[V]$ is open in $X$ and $f : X \to Y$ is continuous. $\square$

I think this proof is probably incorrect as I have not used the definition of coherence in this proof, (my error is probably a set-theoretic issue).

Is this proof correct or incorrect? If it is incorrect, could someone provide a proof of the definition of the quoted property?


Hence $f^{-1}[V]$ is open in $X$

That's the problematic statement. Normally each $f^{-1}_{|B}[V]$ is open in $B$, not in $X$. Thus the union does not have to be open in $X$. And none of your set equalities implies that.

It is open precisely because the topology is coherent with $\mathcal{B}$. The argument is very simple:

$$f^{-1}[V]\cap B=f^{-1}_{|B}[V]$$

The right side is open in $B$ for every $B\in\mathcal{B}$ and thus (by the definition of coherent topology) $f^{-1}[V]$ is open in $X$.

Note that you don't need other set-theoretic equalities you've shown.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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