Identity Mapping to Coarser Topology is Continuous

Theorem

Let $S$ be a set.

Let $\tau_1$ and $\tau_2$ be topologies on $S$.

That is, let $T_1 = \struct {S, \tau_1}$ and $T_2 = \struct {S, \tau_2}$ be topological spaces.

Let $I_S: \struct {S, \tau_1} \to \struct {S, \tau_2}$ denote the identity mapping on $S$:

$\forall x \in S: \map {I_S} x = x$

Then:

$I_S: T_1 \to T_2$ is a continuous mapping

if and only if:

$\tau_2$ is coarser than $\tau_1$.

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Proof 1

But $\tau_2 \subseteq \tau_1$ is the definition of $\tau_2$ being coarser than $\tau_1$.

$\blacksquare$

Proof 2

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Sufficient Condition

Suppose:

$I_S: T_1 \to T_2$ is a continuous mapping.

Then, by definition of everywhere continuous mapping between $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$:

$U_2 \in \tau_2 \implies I_S^{-1} \sqbrk {U_2} \in \tau_1$

where $I_S^{-1} \sqbrk {U_2}$ denotes the preimage of $U$ under $I_S$.

By Preimage of Subset under Identity Mapping, we have:

$U_2 \in \tau_2 \implies U_2 \in \tau_1$

So:

$\tau_2 \subseteq \tau_1$

Hence, by definition of coarser topology:

$\tau_2$ is coarser than $\tau_1$.

$\Box$

Necessary Condition

Suppose:

$\tau_2$ is coarser than $\tau_1$.

That is:

$U_2 \in \tau_2 \implies U_2 \in \tau_1$

By Preimage of Subset under Identity Mapping, we have:

$U_2 \in \tau_2 \implies I_S^{-1} \sqbrk {U_2} \in \tau_1$

where $I_S^{-1} \sqbrk {U_2}$ denotes the preimage of $U_2$ under $I_S$.

So the mapping $I_S$ is by definition, everywhere continuous mapping between $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$.

Hence:

$I_S: T_1 \to T_2$ is a continuous mapping.

$\Box$

Hence the result.

$\blacksquare$