Characteristic properties: a map into a space with the initial topology is continuous iff every composite with the defining family is, a map out of a space with the final topology is continuous iff every composite with the defining family is, and the two topologies are respectively the coarsest and the finest making that family continuous

Let $X$ be a set and let $I$ be an index set.

Initial. Let $(Y_i, \mathcal{T}_i)$ be spaces and $f_i : X \to Y_i$ functions, and give $X$ the initial topology $\mathcal{T}^{\mathrm{in}}$ of the family $(f_i)_{i \in I}$ (The initial topology of a family of maps into spaces and the final topology of a family of maps out of spaces, and the subspace topology as the model initial topology). Then:

1. Every $f_i$ is continuous for $\mathcal{T}^{\mathrm{in}}$, and $\mathcal{T}^{\mathrm{in}}$ is the coarsest topology on $X$ with that property: every topology on $X$ making all the $f_i$ continuous contains $\mathcal{T}^{\mathrm{in}}$.
2. Characteristic property. For every space $Z$ and every function $h : Z \to X$, $h \text{ is continuous } \iff f_i \circ h \text{ is continuous for every } i \in I .$

Final. Let $(Z_i, \mathcal{S}_i)$ be spaces and $g_i : Z_i \to X$ functions, and give $X$ the final topology $\mathcal{T}^{\mathrm{fin}}$ of the family $(g_i)_{i \in I}$. Then:

3. Every $g_i$ is continuous for $\mathcal{T}^{\mathrm{fin}}$, and $\mathcal{T}^{\mathrm{fin}}$ is the finest topology on $X$ with that property: every topology on $X$ making all the $g_i$ continuous is contained in $\mathcal{T}^{\mathrm{fin}}$.
4. Characteristic property. For every space $W$ and every function $k : X \to W$, $k \text{ is continuous } \iff k \circ g_i \text{ is continuous for every } i \in I .$

Claims 2 and 4 determine their topologies: a topology on $X$ satisfying claim 2 for every $Z$ and $h$ must equal $\mathcal{T}^{\mathrm{in}}$, and likewise for claim 4, by the argument recorded in the remarks.

• The characteristic property pins the topology down. Suppose two topologies $\mathcal{T}_1$ and $\mathcal{T}_2$ on $X$ both satisfy claim 2 for every space $Z$ and every function $h$. Apply claim 2 for $\mathcal{T}_1$ to $Z = (X,\mathcal{T}_2)$ and $h = \mathrm{id}$: the composites $f_i$ are continuous on $(X,\mathcal{T}_2)$ by claim 2 for $\mathcal{T}_2$ applied to the identity of $(X,\mathcal{T}_2)$, so the identity $(X,\mathcal{T}_2) \to (X,\mathcal{T}_1)$ is continuous, that is $\mathcal{T}_1 \subseteq \mathcal{T}_2$. Exchanging the roles gives equality. The same argument with the arrows reversed does claim 4.

• Only continuity of the composites is tested, never their openness. Claim 2 says nothing about whether $h$ is open or closed, and claim 4 says nothing about $k$; the constructions below acquire such properties one at a time and each is proved where it is used.

• The one-element family is not a degenerate case but the main one. The subspace topology is the initial topology of a single inclusion and the quotient topology is the final topology of a single surjection (The initial topology of a family of maps into spaces and the final topology of a family of maps out of spaces, and the subspace topology as the model initial topology), so claims 2 and 4 with $I$ a one-element set already carry the characteristic properties of both.