How Does Mapping Affect Induced Topologies in Non-Injective Cases?

[jostpuur]

If Y is a topological space, and f:X->Y is some mapping, we can always give X the induced topology, which consists of the preimages of the open sets in Y. However, this doesn't seem to be a good way if f is not injective. Aren't the open sets in X being left unnecessarily big?

Suppose we have the usual topology in S^1 and the natural mapping $f:\mathbb{R}\to S^1$, $f(x)=(\cos(x),\sin(x))$. Is the induced topology from S^1 the usual topology of R? It doesn't look like that to me. All the open sets are periodic.

I though I could define a more reasonable topology to X like this: $V\subset X$ is open $\Longleftrightarrow$ $f(V)\subset Y$ is open. Now also smaller sets in X can be open. But I noticed that I don't know how to prove that the finite intersection of such "open sets" would still be open, because $f(\cap V_i)=\cap f(V_i)$ doesn't work, and now I'm not sure if this way of trying to define topology works at all.

So... How do you get the topology from Y to the X so that open sets are not forced to be too big?

 

[JasonRox]

If
Suppose we have the usual topology in S^1 and the natural mapping $f:\mathbb{R}\to S^1$, $f(x)=(\cos(x),\sin(x))$. Is the induced topology from S^1 the usual topology of R? It doesn't look like that to me. All the open sets are periodic.

Must it be R? No.

Try looking at the Quotient Topology section in your textbook.

 

[jostpuur]

Try looking at the Quotient Topology section in your textbook.

If I have topology in X, and an equivalence relation there, then I can define a quotient topology in $X/\sim$, but $X/\sim$ is smaller set than X. I don't see how this helps me here, because I'm now trying to get a topology from a smaller set to a bigger one. More precisely, I want a topology from Y to X, when f:X->Y is not injective.

Or did you mean, that we get the topology from $X/\sim$ to X?

 

[jostpuur]

A quick thought! If I say that the collection of sets $\{V\subset X\;|\;f(V)\subset Y\;\textrm{open}\}$ is the subbasis of the topology in X, I guess then everything is fine?

No! That is not what I want. The topology becomes well defined, but it is not of the correct kind. For example in the case of $f:\mathbb{R}\to S^1$, a set $]0,1/2]\cup [1/2+2\pi, 1+2\pi[$ would become open.

 

[JasonRox]

I hope you found my monograph entertaining.

Makes me think, so I enjoy it.

I see what you're saying now. But what you did doesn't seem to go anywhere because you're letting a set be a subbasis. Which really doesn't do anything special because it's obviously going to create some kind of topology.

You might now want to ask yourself the following questions...

How does this topology on X dependent on the map f?

Is it uniquely determined by f?

What kind of structure might it have depending on Y?

 

[jostpuur]

I hope you found my monograph entertaining.

Makes me think, so I enjoy it.

And now it continues, because I edited this temporary ending away.

I have two different ways of defining a topology to X, out of topology of Y and a mapping f:X->Y. One is the usual inducing, and the other one goes through the subbasis. At the moment I have only one requirement for the topology of X. I want to it to be a kind of topology, that if I get the topology from S^1=Y to R=X, I get the usual topology of R. These two ways give something else, so I don't like them.

 

[jostpuur]

I asked about this from one mathematician today, and got an answer. If we have the mapping $f:\mathbb{R}\to S^1$, $f(t)=(\cos(t),\sin(t))$, and the usual topology in $S^1$, there's no way of getting the usual topology to the $\mathbb{R}$ honestly with this mapping only.

Consider for example a sequence 1, 1/2, 1/3, ... Is it supposed to converge towards 0 or towards $2\pi$? There's no way of deciding it using the f only.