example of non-complete lattice homomorphism

The real number line $[-\mathrm{\infty },\mathrm{\infty }]=ℝ\cup \{-\mathrm{\infty },\mathrm{\infty }\}$ is completein its usual ordering of numbers. Furthermore, the meet of a subset $S$ of $ℝ$ isthe infimum of the set $S$.

Now define the map $f:[-\mathrm{\infty },\mathrm{\infty }]\to [-\mathrm{\infty },\mathrm{\infty }]$ as

First notice that if $x\le y$ then $f(x)\le f(y)$, for either $x\le y\le 0$ inwhich case $f(x)=0=f(y)$, or which gives or so $f(x)=1=f(y)$.

In the second place, if $S$ is a finite subset of $ℝ$ then $S$ containsa minimum element $s\in S$. So $f(s)\in f(S)$ and $f(s)\le f(t)$ for all $t\in S$,so $f(\min S)=f(s)=\min f(S)$. Hence $f$ is a lattice homomorphism.

However, $f$ is not a complete lattice homomorphism. To see this let. Then $infS=0$. However,$f(infS)=f(0)=0$ while $inff(S)=inf\{1\}=1$.