Does there exist a continuous function from the unit interval to a circle with "closure"? 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?
 A: 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.
A: 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.
