How to prove that the induced topology is the coarsest and identification topology is the finest topology that keeps the map continuous?

Question

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.

Answer 1

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 $(X, \tau_X)$.

Answer 2

First, let's talk about the induced topology. I'll assume that you know how to verify that this is, in fact, a topology. But I will say that you should keep in mind that the requirements for a topology are a bit stricter than the ones you've written. You must also have that the whole space is open and that the union of any collection of open sets (not just the union of two open sets) is open.

We wish to show that $f$ is continuous when viewed as a map from $(X, f^{-1}(\tau))$ to $(Y, \tau)$. To prove this, we must show that for every $U \in \tau$, we have $f^{-1}(U) \in f^{-1}(\tau)$. But this self-evidently true as part of the very definition of $f^{-1}(\tau)$, since $f^{-1}(\tau)$ is the collection of all subsets of $X$ which can be written as $f^{-1}(U)$ for some $U \in \tau$. So $f^{-1}(\tau)$ definitely makes $f$ continuous.

To show that $f^{-1}(\tau)$ is the coarsest topology making $f$ continuous, we must show that it is a subset of any topology at all on $X$ that makes $f$ continuous. Now suppose we have some topology $\pi$ on $X$ such that $f$ is continuous with respect to this topology. To show that $f^{-1}(\tau) \subseteq \tau$, we must consider an arbitrary element $V \in f^{-1}(\tau)$ and demonstrate that $V \in \pi$. To do this, we may take some $U \in \tau$ such that $V = f^{-1}(U)$. Then since $f$ is continuous w.r.t. $\pi$, we must have $f^{-1}(U) \in \pi$. This demonstrates that $f^{-1}(\tau) \subseteq \pi$. Thus, $f^{-1}(\tau)$ is in fact the coarsest topology making $f$ continuous, as claimed.

As for the identified topology, we can show that it makes $p$ continuous quite straightforwardly. For suppose that we have some $A \in p(\tau)$. Then by definition, $p^{-1}(A) \in \tau$. This completes the proof.

To show that it is the finest such topology, we must show that for any topology $\pi$ on $Y$ such that $\pi$ makes $p$ continuous, $\pi \subseteq p(\tau)$. To do this, consider such a $\pi$. Consider some $U \in \pi$. Then since $p$ is continuous w.r.t. $\pi$, we have $p^{-1}(U) \in \tau$. Therefore, $U \in p(\tau)$. Thus, we have $\pi \subseteq p(\tau)$. Then $p(\tau)$ is in fact the finest topology making $p$ continuous.