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Unit interval $I$ and $S^1$ have different topology - if one identifies the opposite ends of $I$ to make $S^1$, lots of points which were "far" in $I$ (in terms of a number of shared open sets), become "near" in $S^1$.

Is such a function $f: I \to S^1$ continuous?

Following the definition of continuous function: function $f: X \to Y$ is continuous if every open subset $S_Y \subseteq Y$ has an open preimage $S_X \subseteq X$, it seems that the place where opposite ends of the interval $I$ were "joined" has open preimages in $I$.

These are just unions of open sets, which are in turn open sets, so the answer to the question whether $f$ is continuous seems to be yes.

But I suspect that there is a catch, and I don't see some elementary contradiction. Overall, can continuous functions between topological spaces disrupt the topology of the domain, and if not, how to show that?

Entsy
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    Without making further hypotheses (such as injectivity or surjectivity), you can find lots of continuous functions. For example, any constant function is continuous. The unit circle and the (open? closed?) unit interval are note homeomorphic, which means that there is no continuous bijection with a continuous inverse between the two spaces. Other continuous functions are possible. – Xander Henderson Jul 18 '18 at 14:44
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    You can continuously transform a line segment to the circle, the two endpoints will get closer and closer to each other, colliding in the end. However, you can't transform it back. – Berci Jul 18 '18 at 14:45
  • How one shows in this case that we can't find $g: S^1 \to I$ such that $g$ is continuous? For example I've squashed $S^1$ from top and bottom to $I$, and again I'm looking for preimages of open sets in $I$ - again these seem to be unions of open sets in $S^1$, but this time something went wrong for sure :) As you might guess I'm only starting with algebraic topology and such basic questions are a bit unintuitive to me. – Entsy Jul 19 '18 at 11:56

2 Answers2

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Yes, the function $x \mapsto e^{2\pi ix}$ is continuous. As you mention, it is a quotient map that identifies the endpoints of the interval $[0,1]$.

To your second question, of course a continuous map need not preserve the topology of the domain. Constant functions are one easy example. A continuous (surjective) function identifies every open set in the range space with an open set in the domain, but obviously the converse is not true unless your function is a homeomorphism.

mheldman
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Indeed continuous functions alone do not need to "play well" with the topologies of their domain.

However, in your case, this is an example of a quotient. The issue of continuity of a quotient map is usually circumvented by the notion of a quotient topology. Indeed a map $q:X \to X/{\sim}$ has the final topology with respect to the quotient map, meaning the finest topology making the quotient map continuous.

In this case, you are asking about the quotient $Y:=[0,1]/{\sim}$ where $0 \sim 1$. In this case, due to the map having the finest topology, it follows that whenever you have a map $f:[0,1] \to X$ where $f(0)=f(1)$, there exists a unique map $\tilde{f}: Y \to X$ so that $f=\tilde{f} \circ q$. This is the universal property of the quotient map.

Indeed, this is how we show that "quotients" are rigorously homeomorphic to well-known things.

For example, take $\mathrm{exp}:[0,1] \to S^1 \subset \mathbb C$, where $x \mapsto e^{2 \pi ix}$.

Then, since $\mathrm{exp}(0)=\mathrm{exp}(1)$, it follows from the previous discussion that there exists a map $\widetilde{\mathrm{exp}}:Y \to S^1$, but this is a homeomorphism since it is a ontinuous bijection from a compact space to a hausdorff space.

Since $Y$ is evidently homeomorphic to $S^1$, it is customary to just say that the quotient is $S^1$ and dthat $\mathrm{exp}$ is the quotient map up to composition with a homeomorphism from a topological point of view.

Andres Mejia
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  • Andres, thank you for mentioning the universal property of the quotinent map, I wasn't aware of it. One more question: what about a continuous function $g: S^1 \to I$, does it exist? How one shows whether it do/doesn't exist, using the basic definition of continuous function? – Entsy Jul 20 '18 at 09:44
  • You need further conditions. Of course you can have the constant function. – Andres Mejia Jul 21 '18 at 17:04
  • If you only care "up to homotopy," then classes of maps $[S^1,I]=\pi_1(I)=0$ since $I$ is contractible. So, up to homotopy there is only the constant function. – Andres Mejia Jul 21 '18 at 17:05