# Lattice homomorphism induced by topology homeomorphism?

My book mentions this. However, I am not seeing why 1. and 2. are true. I guess part of the problem is that the lattice (homomorphic) map induced by the continuous topology map isn't even defined on the book (it isn't on Section 1.4.9 either). Can anyone tell me how is it usually defined, and maybe help in the subsequent statements too?

• This is essentially the definition of continuity: if $U$ is open in $Y$, then $f^{-1}(U)$ is open in $X$. So the lattice map can just be defined as $f^{-1}$. It's not hard to see that this is indeed a lattice homomorphism. Knowing what the lattice map is, can you then prove 1) and 2)? – user98602 Aug 9 '14 at 19:24
• So are you saying that the "lattice map" they refer to is from $\tau_{2}$ to $\tau_{1}$ and not the other way? as that's the only map I can see defining from $f^{-1}$ If so I would see that map actually satisfying 1. and 2. ! – guillefix Aug 9 '14 at 19:31
• Yup, check closely on lines 4-5: "such a map induces a homomorphism from the lattice $\tau_2$ into the lattice $\tau_1$". – user98602 Aug 9 '14 at 19:34
• I'm probably being dense, but if $X=Y=\{0,1\}$, $\tau_1$ is the discrete topology, $\tau_2$ is the trivial topology, and $f\colon X\to Y$ is the identity map, then $f$ is continuous one-to-one, but the induced map $f_*\colon\tau_2\to\tau_1$ is a map from a $2$-element set to a $4$-element set, so can't be surjective. Am I missing something? – Dan Rust Aug 9 '14 at 19:52
• @DanielRust No, guillefix, Daniel is right. Continuity only requires that the inverse image of every open set is open; it does not require the converse. Statement 1 in the book is wrong. – Andreas Blass Aug 9 '14 at 20:10

As discussed in the comments, the lattice map is the inverse set map, $f^{-1}(A) = \{ x \in X | f(x) \in A \}$ from $\tau_{2}$ to $\tau_{1}$. As also pointed out by @DanielRust in the comments, the first statement is false as he gives a valid counter-example.
Take two open sets in $\tau_{2}$, $U$ and $V$. Take $f^{-1}(U) = f^{-1}(V) = A \in \tau_{2}$ (as $f$ is continuous) Now, from the definition of $f^{-1}$, $f(A) = \{f(x) | x \in A \} \subset U, V$. Now, take $y \in U$. Because $f$ is surjective, $y = f(x')$ for some $x' \in X$. But this implies that $x' \in f^{-1}(U)$ and thus that $y \in f(A)$. Therefore, $U \subset f(A)$, which can be similarly shown for $V$. This, at once, shows that $U = f(A) = V$, and that $f^{-1}$ is surjective.