This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
In §Continuous Functions, for a continuous function \(f:X \rightarrow Y\) we defined the function \(f^\mathcal{T}:\mathcal{T}_Y\rightarrow\mathcal{T}_X\) via the formula
\[f^\mathcal{T}(V)=f^{-1}(V)\]But conversely, for an arbitrary function \(f:X\rightarrow Y\), the function that takes an arbitrary subset of \(Y\) and assigns its preimage under \(f\)
\[f^\text{pre}:\mathcal{P}(Y)\rightarrow\mathcal{P}(X); \qquad V\mapsto f^{-1}(V)\]is always well-defined, and its restriction to the subset \(\mathcal{T}_Y\) of \(\mathcal{P}(Y)\) is exactly \(f^\mathcal{T}\). In this case, the image of the function \(f^\text{pre}\vert_{\mathcal{T}(Y)}\) landing in \(\mathcal{T}_X\subseteq \mathcal{P}(X)\) is precisely the condition for \(f\) to be continuous.
In particular, if we fix an arbitrary set \(X\) with the discrete topology \(\mathcal{T}_1\) and the trivial topology \(\mathcal{T}_2\) defined on it, any function from the topological space \((X, \mathcal{T}_1)\) to an arbitrary topological space \((Y, \mathcal{T})\) is continuous, and any function from an arbitrary topological space \((Y,\mathcal{T})\) to \((X, \mathcal{T}_2)\) is always continuous.
Initial topology
In this post we examine topological structures defined by similar properties.
Definition 1 Let a set \(X\) and a family of topological spaces \((Y_i,\mathcal{T}_i)_{i\in I}\) be given, and for each \(i\) let a function \(f_i:X\rightarrow Y_i\) be given. The weakest topology on the set \(X\) making all the functions \(f_i\) continuous is called the initial topology defined by the \(f_i\).
For topologies \((\mathcal{T}_j)_{j\in J}\) defined on \(X\), suppose all the functions \(f_i\) from the topological space \((X, \mathcal{T}_j)\) to \((Y_i,\mathcal{T}_i)\) are continuous. Then \(\mathcal{T}=\bigcap_{j\in J}\mathcal{T}_j\) defines a topology on \(X\), and we can show that all the functions \(f_i\) from this topological space \((X,\mathcal{T})\) to \((Y_i,\mathcal{T}_i)\) are continuous. Moreover, since any function with the discrete topology as its domain is always continuous, a topology satisfying this condition always exists, and thus it follows immediately that the initial topology always exists. As is always the case with this kind of argument, this is an impeccable argument for showing the existence of the initial topology, but it is of little help in seeing what the initial topology actually looks like. Therefore we need to examine the situation a little more concretely.
For the function \(f_i\) to be continuous, \(f_i^{-1}(U_i)\) must be an open set in \(X\) for any open set \(U_i\) of \(Y_i\); therefore the initial topology we are going to define must contain all elements of the form \(f_i^{-1}(U_i)\). On the other hand, a topological space containing these elements must also contain their finite intersections and arbitrary unions. Hence we can prove the following proposition.
Proposition 2 The initial topology of Definition 1 is exactly the topology generated by taking the following set
\[\mathcal{S}=\{f_i^{-1}(U_i)\mid \text{$U_i$ open in $Y_i$}\}\]as a subbase.
Proof
Write the initial topology as \(\mathcal{T}_\ini\), and the topology generated by taking \(\mathcal{S}\) as a subbase as \(\mathcal{T}\). Since \(\mathcal{T}\) makes all the \(f_i\) continuous by definition, \(\mathcal{T}_\ini\) is weaker than \(\mathcal{T}\). Thus it suffices to show that \(\mathcal{T}\) is weaker than \(\mathcal{T}_\ini\), which is immediate because \(\mathcal{T}\) is the weakest topology containing \(\mathcal{S}\).
Then the initial topology has the following kind of universal property.
Proposition 3 In the situation of Definition 1, suppose furthermore that a topological space \(Z\) and a map \(g:Z\rightarrow X\) are given. Then \(g\) is continuous if and only if each \(f_i\circ g\) is continuous.
Proof
If \(g\) is continuous then \(f_i\circ g\) is trivially continuous as a composition of continuous functions. Thus we only need to show the converse.
Suppose each function \(f_i\circ g\) is continuous. For any open subset \(U\) of \(X\), by Proposition 2 there exist \(U_j\) such that
\[U=\bigcap_{j=1}^n f_j^{-1}(U_j)\]holds. Therefore
\[g^{-1}(U)=g^{-1}\left(\bigcap f_j^{-1}(U_j)\right)=\bigcap_{j=1}^n(f_j\circ g)^{-1}(U_j)\]and by assumption \((f_j\circ g)^{-1}(U_j)\) is an open set, so \(g^{-1}(U)\) must also be an open set. That is, \(g\) is continuous.
Final topology
Definition 4 Let a set \(X\) and a family of topological spaces \((Y_i,\mathcal{T}_i)_{i\in I}\) be given, and for each \(i\) let a function \(f_i:Y_i\rightarrow X\) be given. The strongest topology on \(X\) making all the \(f_i\) continuous is called the final topology induced by the \(f_i\).
If the trivial topology is given on \(X\), any function into \(X\) is always continuous. But in general the union of topologies is not a topology, so unlike the initial topology, the existence proof relies heavily on the following proposition.
Proposition 5 The topology in Definition 4 is defined as the following set
\[\mathcal{T}_\fin=\{U\subseteq X\mid f^{-1}_i(U)\text{ is open in $Y_i$ for all $i$}\}\]Proof
That the given collection \(\mathcal{T}_\fin\) is actually a topology can be easily verified. Thus it suffices to show that \(\mathcal{T}_\fin\) satisfies all the conditions of Definition 4.
First, for any \(U\in\mathcal{T}_\fin\) and any \(i\), that \(f_i^{-1}(U)\) is open in \(Y_i\) is clear from the definition of \(\mathcal{T}_\fin\). On the other hand, suppose another topology \(\mathcal{T}\) on \(X\) satisfying the given condition is given. Then for any \(U\in\mathcal{T}\), \(f^{-1}_i(U)\) must be open in \(Y_i\). Therefore, by the definition of \(\mathcal{T}_\fin\), we have \(U\in\mathcal{T}_\fin\), and hence \(\mathcal{T}_\fin\) is stronger than \(\mathcal{T}\).
Likewise, the final topology also satisfies the following universal property similar to that of the initial topology.
Proposition 6 In the situation of Definition 4, suppose furthermore that a topological space \(Z\) and a map \(g:X\rightarrow Z\) are given. Then \(g\) is continuous if and only if each \(g\circ f_i\) is continuous.
Proof
If \(g\) is continuous then \(g\circ f_i\) is trivially continuous as a composition of continuous functions. Thus we only need to show the converse.
Suppose each function \(g\circ f_i\) is continuous. Then for any open set \(U\subseteq Z\), the following sets
\[(g\circ f_i)^{-1}(U)=f_i^{-1}(g^{-1}(U))\]are each open in \(Y_i\). But by Proposition 5, this is equivalent to saying that \(g^{-1}(U)\) is an open set in \(X\), and therefore \(g\) is continuous.
References
[Bou] N. Bourbaki, General Topology. Elements of mathematics. Springer, 1995.
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