Final Topology with respect to Mapping

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Theorem

Let $\struct {X, \tau_X}$ 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:

$\tau_Y = \set {U \subseteq Y : f^{-1} \sqbrk U \in \tau_X}$

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

Further, the following holds true:

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

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

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