Final Topology with respect to Mapping

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Theorem

Let $\left({X, \tau_X}\right)$ be a topological space.

Let $Y$ be a set.

Let $f: X \to Y$ be a mapping.

Let $\tau_Y$ be the final topology on $Y$ with respect to $f$.

Then:

$\tau_Y = \left\{ U \subseteq Y : f^{-1} \left({U}\right) \in \tau_X \right\}$

Observe that the set on the right-hand side of the equality is sometimes denoted $f\tau$.

Further, the following holds true:

  1. $\forall \left(Z, \tau_Z \right)$ topological space $: \forall g: \left({Y, f\tau}\right) \to \left({Z, \tau_Z}\right)$ mapping $: g$ is continuous $\iff g \circ f$ is continuous (where $f$ is defined from $\left({X, \tau_X}\right)$ to $\left(Y, f\tau \right)$).
  2. $\forall \tau_Y$ topology in $Y : \forall \left(Z, \tau_Z \right)$ topological space, the following are equivalent:
    1. $\tau_Y = f\tau$
    2. $\forall g: \left({Y, \tau_Y}\right) \to \left({Z, \tau_Z}\right) : g$ is continuous $\iff g \circ f$ is continuous (where $f$ is defined from $\left({X, \tau_X}\right)$ to $\left(Y, \tau_Y \right)$).

That is, the topology $f\tau$ is characterised by the fact that it is the only one which makes all mappings defined from it to any topological space continuous.


Proof