Let $f: (X, \sigma) \rightarrow (Y, \tau)$ be continuous and let $a \sim b$ if $f(a) = f(b)$. Noting $\sim$ is an equivalence relation, let $\pi(x)$ map $x$ to its equivalence class. Noting $\pi: X \rightarrow X/\sim$ is onto, give $X/\sim$ the identification topology $\upsilon$ determined by $\pi$. Since $\pi(a) = \pi(b)$ implies $f(a) = f(b)$, $f$ induces a continuous function $g: X/\sim\, \rightarrow Y$ such that $f = g\pi$ (see Theorem 8.2, Chapter 3 of Mendelson's Introduction to Topology). We have
$g(\pi(a)) = g(\pi(b)) \Rightarrow f(a) = f(b) \Rightarrow \pi(a) = \pi(b)$,
and so $g$ is one-to-one. Since $g$ is continuous, $g^{-1}(\tau) \subset \upsilon$.
Confusion:
According to Mendelson's Introduction to Topology (page 103), since $g$ is one-to-one, $g^{-1}(\tau) \subset \upsilon$ is equivalent to $\tau \subset g(\upsilon)$, and therefore, since $\tau$ is any topology on $Y$, the topology 'carried over' to $Y$ by $g$ is the weakest topology such that $f$ is continuous.
It is not clear to me that either of these statements make sense unless $g$ is onto. However, even if $g$ is onto, for any $T \in \tau$, $g^{-1}(T)$ is open, and so the identification topology on $Y$ determined by $g$ contains $T$. Therefore this identification topology ought to be strongest topology on $Y$ such that $f$ is continuous. Here I am assuming that the topology 'carried over' to $Y$ by $g$ is the identification topology.
I am already a bit suspicious since Mendelson never specified that $f$ must be continuous in order to induce $g$ (I added this requirement myself).
Questions:
- Is it correct that $g^{-1}(\tau) \subset \upsilon \iff \tau \subset g(\upsilon)$?
- Am I correct in assuming that the topology 'carried over' to $Y$ by $g$ is the identification topology?
- Is the topology 'carried over' to $Y$ by $g$ is the weakest topology such that $f$ is continuous?