ThmDex – An index of mathematical definitions, results, and conjectures.
Result R298 on D55: Continuous map
Identity map is continuous if topology on domain set is equal or stronger
Formulation 0
Let $T_0 = (X, \mathcal{T}_0)$ and $T_1 = (X, \mathcal{T}_1)$ each be a D1106: Topological space.
Let $I : X \to X$ be an D440: Identity map on $X$.
Then
(1) If $\mathcal{T}_1 \subseteq \mathcal{T}_0$, then $I$ is a D55: Continuous map with respect to $T_0$ and $T_1$
(2) If $T_0 = T_1$, then $I$ is a D55: Continuous map with respect to $T_0$
Proofs
Proof 0
Let $T_0 = (X, \mathcal{T}_0)$ and $T_1 = (X, \mathcal{T}_1)$ each be a D1106: Topological space.
Let $I : X \to X$ be an D440: Identity map on $X$.
Suppose that $\mathcal{T}_1$ is coarser than $\mathcal{T}_0$. If $U \in \mathcal{T}_1$, then $I^{-1}(U) = U \in \mathcal{T}_1 \subseteq \mathcal{T}_0$, so $U$ is in $\mathcal{T}_1$ by transitivity of the subset relation. Since $U$ was an arbitrary open set in $T_1$, we may use result R324: Continuity characterised by preimages of open sets to conclude that $f$ is continuous with respect to $T_0$ and $T_1$. If $T_0 = T_1$, then $\mathcal{T}_0 = \mathcal{T}_1$ and $\mathcal{T}_1 \subseteq \mathcal{T}_0$ holds due to R125: Subset relation is reflexive. $\square$