We know that if $X$ locally compact space and $f:X\to Y$ is a continuous mapping then $f(X)$ may not be locally compact. But it we require $f$ to be an open map then the image $f(X)$ will be locally compact!
And the classical example is the following: we have to take the mapping $i:\mathbb{Q_d}\to \mathbb{Q}$ where $i(x)=x$ and $\mathbb{Q}_d$ is the rationals with discrete topology and $\mathbb{Q}$ rationals under usual topology. But $\mathbb{Q}_d$ is locally compact however $\mathbb{Q}$ is not locally compact.
However, I came up with a bit different solution: Consider the function $f:\mathbb{R}\to \mathbb{R}^{\omega}$ given by rule $f(x)=(x,0,0,\dots)$ and easy to check that $f$ is continuous, $\mathbb{R}$ locally compact but $\mathbb{R}^{\omega}$ is not locally compact.
Is my example correct?
Best Answer
Any space is the continuous (bijective) image of a locally compact and locally connected (as a bonus) space (namely, the discrete topology on the same set). You use that in the first (correct) example.
Your second example does not work as the image is just a homeomorphic copy of $\Bbb R$ so it is locally compact. You don't get all of $\Bbb R^\omega$.