Prove that the induced topology is the coarsest and identification topology is the finest topology that keeps the map continuous

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I am reading maps between topological space from Isham, Chris J. Modern differential geometry for physicists. Vol. 61. World Scientific, 1999.. Here he defines the induced topology and the identification topology in the following way

If $(Y,\tau)$ is a topological space and $f$ is a map from $X$ to $Y$
then the induced topology on $X$ is defined to be $$
f^{-1}(\tau):=\{f^{-1}(O)|O \in \tau\} $$ The key property of the
induced topology is that it is the coarsest topology such that $f$ is
continuous

Another important example arises when $(Y,\tau)$ is a topological
space and there is a surjective map $p: Y \to X$. The identification
topology on $X$ is defined as
$$
p(\tau) := \{A\subset X | p^{-1}(A) \in \tau\}
$$ The key property of this topology is that it is the
finest one on $X$ such that $p$ is continuous.

I want to prove that

  • the induced topology is the coarsest topology such that $f$ is continuous
  • the identification topology is the finest topology such that $p$ is continuous

I know that a map $f:(W,\tau) \to (V,\tau')$ is a continuous map if for all $O \in\tau', f^{-1}(O) \in \tau$.

I also know that in general, any maps between two sets ($f:A\to B$) have the property
$$
f^{-1}(A\cap B) = f^{-1}(A) \cap f^{-1}(B) \\
f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)
$$
This comes handy to prove that both of the induced and identification topologies are topologies.

But I don't know where to start with the proofs.

Best Answer

It's an exercise in definitions:

Recall that given $f:X \to (Y, \tau_Y)$, we define $\tau_f = \{f^{-1}[O]\mid O \in \tau_Y\}$, which is indeed a topology on $X$, from the properties of inverse images you name, among other things.

It's trivial that $f$ is continuous when $X$ is given the induced topology $\tau_f$ wrt $(Y,\tau_Y)$: let $O \in \tau_Y$ be open. By definition of the induced topology $\tau_f$, $f^{-1}[O] \in \tau_f$. So $f$ is continuous as a map $(X,\tau_f) \to (Y, \tau_Y)$.

In fact, if $\tau$ is any topology on $X$ such that $f:(X,\tau) \to (Y, \tau_Y)$ is continuous, when $O \in \tau_f$, we know $O=f^{-1}[O']$ for some $O' \in \tau_Y$, and as $f$ is assumed to be continuous by definition of continuity of $f$, $f^{-1}[O'] \in \tau$, so $O \in \tau$ and $\tau_f \subseteq \tau$. So $\tau_f$ is the coarsest among all topologies that makes $f$ continuous with codomain $(Y, \tau_Y)$.

Now to the identification topology (aka as quotient topology or final topology), when $(X, \tau_X)$ is pre-given and $\tau_i:= \{O \subseteq Y: f^{-1}[O] \in \tau_X\}$ is the identification topology.

Now, if $\tau$ is any topology on $Y$ such that $f:(X, \tau_X) \to (Y, \tau)$ is continuous.

Let $O \in \tau$ arbitrary. Then, as $f$ is continuous, by definition, $f^{-1}[O] \in \tau_X$. But this means that $O$ obeys the defining property for $\tau_i$, so $O \in \tau_i$. And as $O$ is arbitrary, $\tau \subseteq \tau_i$, so the identification topology is the finest (largest) topology among all topologies on $Y$ that make $f$ continuous with domain $(X, \tau_X)$.

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