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?(adsbygoogle = window.adsbygoogle || []).push({});

Suppose we have the usual topology in S^1 and the natural mapping [itex]f:\mathbb{R}\to S^1[/itex], [itex]f(x)=(\cos(x),\sin(x))[/itex]. 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: [itex]V\subset X[/itex] is open [itex]\Longleftrightarrow[/itex] [itex]f(V)\subset Y[/itex] 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 [itex]f(\cap V_i)=\cap f(V_i)[/itex] 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?

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# Different induced topologies

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