0
$\begingroup$

Munkres Topology. In Section 71, coherent is "if" but on Wikipedia (Coherent topology), it's "if and only if"

enter image description here

This can be seen in Example 1 of Section 83

enter image description here

We suppose $D \cap A_{\alpha}$ is closed in $A_{\alpha}$ and then must show $D$ is closed in $X$. I don't think we also suppose $D$ is closed in $X$ and then show that the $D \cap A_{\alpha}$'s are closed.

But in the definition of subgraph (also in Section 83), coherent is "if and only if" (I am aware of the errata by Barbara and Jim Munkres for this definition but irrelevant I think).

enter image description here

I was expecting to see that we suppose $D \cap A_{\beta}$ is closed in $A_{\beta}$ and then must show $D$ is closed in $Y$, but we actually also show We suppose $D$ is closed in $X$ and then show that the $D \cap A_{\alpha}$'s are closed.

What's going on?

My guess (I came up with one only after typing it all up)

Definitions are "if and only if". If there is no specified topology for a space $Z$, then coherence is just "if" and then "only if" follows because coherence is the definition. If there is a specified topology on $Z$, such as it having the subspace topology of some other space, then we have to show that coherence, a condition to indicated closedness of sets, and hence openness (of those sets' complements) doesn't conflict with with our new definition of closedness given by the subspace topology.

I think I figured it out, but I might as well just submit this since I already typed it up.

$\endgroup$
0
$\begingroup$

Definitions, by their nature, are "if and only if". This leads to some authors being lazy when writing them, and they only say "if".

It's a common enough occurrence and something one should always be aware of. The alternative definition supplied by Munkres definitely seems to be one such case.

$\endgroup$
  • $\begingroup$ Is this correct? "If there is a specified topology on $Z$, such as it having the subspace topology of some other space, then we have to show that coherence, a condition to indicated closedness of sets, and hence openness (of those sets' complements) doesn't conflict with with our new definition of closedness given by the subspace topology." $\endgroup$ – user198044 Nov 15 '18 at 7:32
0
$\begingroup$

The notion Munkres gives is the right one in that context. $X$ already has a topology and its subspaces $X_\alpha$ have the subspace topolgoy induced by it, so we already know by definition that $O$ open in $X$ implies $O \cap X_\alpha$ open in $X_\alpha$.

The coherence part comes from the idea that we suppose we are just given the subspace topologies on $X_\alpha$ at the start (and we "forget" the topology on $X$) can we then reconstruct the topology on $X$ from these? In a coherent topology you can: you test for a subset $O$ of $X$ whether indeed each set $O \cap X_\alpha$ is open in the subspace topology in $X_\alpha$, and if this is the case we conclude $O$ must have open in $X$, and we get exactly the original topology on $X$ back.

Because this is the new part in coherence Munkres only states the implication of the second paragraph. The standard pasting lemma's say that any open cover of $X$ or any (locally) finite closed cover of $X$ form a coherent family for $X$.

The coherent topology idea is used to construct the topology on a disjoint sum of spaces, and also in a CW-complex, where we glue subspaces with given topologies together to form new spaces. The graph construction is a special low-dimensional case of a CW-complex.

$\endgroup$

Your Answer

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