Continuous Function From Compact Space to Hausdorff Space is an Identification

Question

Mendelson's Introduction to Topology: Third Edition, Chapter 5, Section 1, Question 6:

Let $f: X \rightarrow Y$ be a continuous mapping of a compact space $X$ onto a Hausdorff space $Y$. Prove that $f$ is an identification.

Answer 1

It suffices to show that if $f^{-1}[C]$ is closed in $X$ for some $C \subseteq Y$ then $C$ is closed in $Y$.

1. $f^{-1}[C]$ is compact (as $X$ is compact).
2. $C=f[f^{-1}[C]]$ because $f$ is onto (i.e. surjective).
3. As $f$ is assumed to be continuous, $C$ is compact.
4. As $Y$ is Hausdorff, $C$ is closed in $Y$.

QED. See how all assumptions are used..

Answer 2

Let both $X$ and $Y$ be the real interval $[0,1]$ and let $f(x) = 0$ for all $x$ in $X$. Then $X$ and $Y$ are both compact and both Hausdorff. A constant function is continuous, but $f$ is not surjective so it is not an identification.

From this, we realise that we are supposed to assume that $f$ is surjective, so then it only remains to show that any subset of $Y$ with open preimage is open. (This part is fine)

But this seems to come up a lot... sometimes the notation $f:X \rightarrow Y$ means that $f$ is mapping into (possibly only a subset of) $Y$, while at other times it seems to be implicitly "understood" that $f$ is surjective. Frustrating...