Prob. 3, Sec. 29, in Munkres' TOPOLOGY, 2nd ed: The image of a locally compact topological space under a continuous (and open) map

Question

Here is Prob. 3, Sec. 29, in the book Topology by James R. Munkres, 2nd edition:

Let $X$ be a locally compact space. If $f \colon X \to Y$ is continuous, does it follow that $f(X)$ is locally compact? What if $f$ is both continuous and open? Justify your answer.

I know that if $f$ is both continuous and open, then $f(X)$ is locally compact. Here is a Math Stack Exchange post containing a proof of how this holds?

What if $f$ is a continuous map that is not open?

I know that there is no such example of a continuous map $f \colon \mathbb{R} \to \mathbb{Q}$, though $\mathbb{R}$ is locally compact and $\mathbb{Q}$ is not, because every such map is constant, as has been discussed here and here.

Can we construct a counter-example of a (surjective) mapping $f \colon \mathbb{R}^n \to \mathbb{R}^\omega$?

After all, $\mathbb{R}^n$ is locally compact, whereas $\mathbb{R}^\omega$ is not, as has been discussed in Example 2, Sec. 29, in Munkres.

Answer 1

We can just take $Y$ equal to your favourite non-locally compact Hausdorff space and $X$ the same set but with the discrete topology on it.

Then taking $f$ to be the identity function, we have that $Y$ is a bijective continuous image of $X$ (what's called a "condensation" of $X$ in a lot of the literature) and $X$ is locally compact, metrisable, etc etc. while $Y$ is not locally compact by our setup.

E.g take $Y=\Bbb Q$ or $\Bbb P$ (the irrationals) or $\Bbb R^\omega$ (product, uniform or box topology; all are not locally compact); pick your favourite.

Indeed for open maps the situation is better: $f:X \to Y$ open and continuous, then let $y \in f[X]$ and let $x \in X$ be such that $f(x)=y$. Then in $X$ there is a compact $C$ such that $U \subseteq C$ for some open neighbourhood of $x$ (definition that Munkres gives on p. 182 of your edition). Then $f[C]$ is compact (by continuity of $f$) and contains the open neighbourhood $f[U]$ of $y$ (as $f$ is open). So $f[X]$ is locally compact.

Answer 2

Take $X=[1,\infty)$, $Y=\mathbb{R}[X]$, the infinite-dimensional vector space of polynomials with real coefficients, endowed with any norm (so that finite-dimensional subspaces are always closed). Let $B_n \subset Y$ be the set of polynomials with all coefficients between $-1$ and $1$ and with degree at most $n$.

We know that there are surjective maps $g_n:[0,1] \rightarrow B_n$ with $g_n(0)=g_n(1)=0$.

Indeed, take a Peano curve, ie a continuous surjection $h=(a,b):[0,1] \rightarrow [0,1]^2$. Then define the continuous surjection $h_n:[0,1]^n \rightarrow [0,1]^{n+1}$ by $h_n(x)=(x_1,\ldots,x_{n-1},a(x_n),b(x_n))$. Consider next the maps $p_n=h_n \circ h_{n-1} \ldots \circ h_1$, that are surjective continuous maps $[0,1] \rightarrow [0,1]^{n+1}$.

Next, modify $p_n$ into $q_n$ such that: $q_n(t)=p_n(2t-1/2)$ for $1 \leq 4t \leq 3$, $q_n(t)=(1-4t)u+4tp_n(0)$ for $0 \leq t \leq 1/4$, and $q_n(t)=(4t-3)u+(4-4t)p_n(1)$ for $3/4 \leq t \leq 1$, with $u=\frac{1}{2}(1,1,1, \ldots,1)$.

Write $q_n=(c_0,\ldots,c_n)$, define $Q_n(t)=\sum_{k=0}^n{c_k(t)X^k}$ and finally $g_n(x)=2Q_n(x)-1 \in B_n$.

Define now $f$ so that $f(x)=ng_n(x-n)$ for $x \in [n,n+1]$ for all $n \geq 1$. $f$ is onto and continuous.

Indeed, $f$ is continuous on all the intervals $(n,n+1)$, $n \geq 1$, and has limits left and right at all integers (equal to $0$), so is continuous at all integers too.

However, $f$ has a locally compact domain but not a locally compact image, by Riesz’s theorem, since $Y$ is an infinite-dimensional normed vector space.