Prove that $f$ is a quotient map.

For $a,b,c,d\in\mathbb{R}$ let $f:[a,b]\rightarrow [c,d]$ be a continuous function such that $c,d\in f([a,b])$. Prove that $f$ is a quotient map.

My attempt:

We need to show that $U$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$, and that $f$ is surjective.

We have one direction since $f$ is continuous. Let $x\in f^{-1}(U)$, where $f^{-1}(U)$ is open in $[a,b]$. Since $f^{-1}(U)$ is open, this implies that there is some basis element $(m,n)$ for $m,n\in[a,b]$ such that $x\in(m,n)\subset f^{-1}(U)$. Therefore we have that $x\in f((m,n))\subset U$. Since $(m,n)$ is connected and $f$ is continuous we know that $f((m,n))$ is connected, and therefore $f((m,n))=(y,z)\subset[c,d]$. This is also a basic open set and therefore $U$ is open.

To show that $f$ is surjective, we want to show that for every $y\in[c,d]$ there exists and $x\in [a,b]$ such that $f(x)=y$. Suppose not. This implies that there is some $y\in[c,d]$ such that no $f(x)\not=y$ for any $x$. Since $f$ is continuous, and $[a,b]$ is connected, the image of a connected set is connected, and since $c,d\in f([a,b])$, if there was some $y\in [c,d]$ that could not be mapped to, we would have a separation of $[c,d]$ which is a contradiction. Therefore $f$ is surjective.

Solution 1:

Careful! I don't think is clear that the image of $(m, n)$ is necessarily an open interval. For instance, let $f(x)=x^2$, and note that $f((-1, 1))=[0,1)$. (This does not satisfies the hypotheses right now; but it is possible to slightly modify it to get into trouble.)

I think it is easy to prove the closed version, i.e. instead of showing that "$U$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$", you can show that "$C$ is closed in $Y$ if and only if $f^{-1}(C)$ is closed in $X$". (If you haven't seen it before, this follows by taking $C=U^c$ essentially.) Note that one direction is clear from continuity; for the other one, if $C \subset Y$ is such that $f^{-1}(C)$ is closed in $X$, then $f^{-1}(C)$ is compact in $X$, as it is a closed subset of the compact set $X$. Therefore, $f(f^{-1}(C))$ is compact in $Y$ (as $f$ is continuous), hence it is closed (as $Y$ is a metric space, of more generally as $Y$ is Hausdorff). But you have proved that $f$ is surjective, hence $f(f^{-1}(C))=C$ is closed, as required.

As a nice remark, this proof is slightly more general: a continuous, surjective map from a compact space to a Hausdorff space is automatically quotient map. (This is really useful!)

Solution 2:

If $U$ is not open in $[c,d]$ then there is a $y\in U$ and a sequence $(y_n)\subseteq [c,d]\setminus U$ such that $(y_n)\to y.$

$[a,b]\setminus f^{-1}(U)$ is closed and bounded, hence compact. Since $f$ is continuous, $f([a,b]\setminus f^{-1}(U))$ is compact hence closed, and it contains $(y_n).$ It follows then that it contains $y$ too, which is a contradiction.

$f$ is surjective by the intermediate value theorem.