I'm trying to construct a continuous surjection from $(0,1]$ onto $[0,1]$, but I'm not getting anywhere. I don't immediately see a contradiction which falsifies the existence of such a function, so my intuition tells me one exists. I feel like an absolute value function would work, but I'm not sure how to arrive at it in the proper way. Thanks for any help.
[Math] A continuous surjective function from $(0,1]$ onto $[0,1]$
continuityfunctionsgeneral-topologyreal-analysis
Related Solutions
I think it is worth giving a counterexample to the assertion : "continuous bijections are homeomorphisms".
The example is a rather simple one : the identity map is always a bijection. If you take two topologies on a set, one which is (strictly) coarser than the other, then the identity map will be continuous precisely in one direction : from the topology with more open sets to the one with less open sets. The other way, if you take a pullback of an open set which is not in the smaller topology, it won't be open in the smaller topology, hence continuity is not possible.
For example, take $\mathbb R$ with the usual topology, and say $\mathbb R$ with indiscrete/discrete topology with the identity map.
This is an example of a continuous bijection which is NOT a homeomorphism. There are conditions under which a continuous bijection is a homeomorphism (compact to Hausdorff), but it is not true in general.
It is true that $f^{-1}$ is closed : in fact, the fact that $f^{-1}$ carries closed sets to closed sets is equivalent to the continuity of $f$. Therefore, we do not expect this fact to be of help to us in the question. I will expand.
When we look at $[0,1]$ and $(0,1)$ as topological spaces, it is with the subspace topology derived from $\mathbb R$. That is, a description of "open" or "closed" in each of these topologies, is given by the intersection of the set with a set that is open or closed in $\mathbb R$ respectively.
For example :
$[0,1]$ is closed and open in the $[0,1]$ topology, because $[0,1] = [0,1] \cap \mathbb R$, and so it is the intersection of $[0,1]$ with a set both open and closed in $\mathbb R$.
Similarly, $(0,1)$ is both open and closed in the subspace topology derived from $\mathbb R$ (in the subspace topology, not the one on $\mathbb R$).
With this in mind, since the map is from $(0,1)$ to $[0,1]$, the fact that $f^{-1}([0,1]) = (0,1)$ is closed is true, and not a contradiction : this is because $f$, as defined as a topological map, is operating with the subspace topologies, not those directly derived from $\mathbb R$. That is the problem with your logic.
Best Answer
Try $f(t) = \sin(1/t)$.
If you want to use your absolute value idea, try making a $V$ shape with the vertex at $(\tfrac12, 0)$, opening upwards to include the points $(0,1)$ and $(1,1)$. (I'll leave the actual function up to you.)