Initial and Final Topology

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In §Continuous Functions we defined, for a continuous function \(f:X \rightarrow Y\), the function \(f^\mathcal{T}:\mathcal{T}_Y\rightarrow\mathcal{T}_X\) by the formula

\[f^\mathcal{T}(V)=f^{-1}(V)\]

Conversely, for any function \(f:X\rightarrow Y\), the map taking an arbitrary subset of \(Y\) to 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 restricting this to the subset \(\mathcal{T}_Y\) of \(\mathcal{P}(Y)\) yields exactly \(f^\mathcal{T}\). In this case, the condition that the image of \(f^\text{pre}\vert_{\mathcal{T}(Y)}\) lies in \(\mathcal{T}_X\subseteq \mathcal{P}(X)\) is precisely the condition that \(f\) is continuous.

In particular, fixing an arbitrary set \(X\) with the discrete topology \(\mathcal{T}_1\) and the trivial topology \(\mathcal{T}_2\) 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 suppose that for each \(i\) a function \(f_i:X\rightarrow Y_i\) is given. The weakest topology on \(X\) making all the functions \(f_i\) continuous is called the initial topology defined by the \(f_i\).

For topologies \((\mathcal{T}_j)_{j\in J}\) on \(X\), suppose that the functions \(f_i\) from the topological space \((X, \mathcal{T}_j)\) to \((Y_i,\mathcal{T}_i)\) are all continuous. Then \(\mathcal{T}=\bigcap_{j\in J}\mathcal{T}_j\) defines a topology on \(X\), and one can show that the functions \(f_i\) from this topological space \((X,\mathcal{T})\) to \((Y_i,\mathcal{T}_i)\) are all continuous. Moreover, since any function whose domain carries the discrete topology is always continuous, such a topology always exists, and therefore the existence of the initial topology is trivial. As is always the case with this kind of argument, this is an impeccable proof of the existence of the initial topology, but it is of little help in seeing what the initial topology actually looks like. Hence we need to examine the situation somewhat more concretely.

For each function \(f_i\) to be continuous, \(f_i^{-1}(U_i)\) must be an open set in \(X\) for every open set \(U_i\) of \(Y_i\), so the initial topology we define must contain all elements of the form \(f_i^{-1}(U_i)\). On the other hand, a topology containing these elements must also contain their finite intersections and arbitrary unions. Therefore we can prove the following proposition.

Proposition 2 The initial topology of Definition 1 is exactly the topology generated by taking the set

\[\mathcal{S}=\{f_i^{-1}(U_i)\mid \text{$U_i$ open in $Y_i$}\}\]

as a subbase.

Proof

Let us 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 trivial because \(\mathcal{T}\) is the weakest topology containing \(\mathcal{S}\).

Then the initial topology has the following universal property.

Proposition 3 In the situation of Definition 1, suppose additionally 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 opposite direction.

Suppose each function \(f_i\circ g\) is continuous. For an arbitrary proper 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 each \((f_j\circ g)^{-1}(U_j)\) 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 suppose that for each \(i\) a function \(f_i:Y_i\rightarrow X\) is given. The strongest topology on \(X\) making all the \(f_i\) continuous is called the final topology defined by the \(f_i\).

If the trivial topology is given on \(X\), then any function into \(X\) is always continuous. But since the union of topologies is not generally a topology, unlike the initial topology, the existence proof depends heavily on the following proposition.

Proposition 5 The topology of Definition 4 is defined by the 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 set \(\mathcal{T}_\fin\) is actually a topology is easily verified. Thus it suffices to show that \(\mathcal{T}_\fin\) satisfies all the conditions of Definition 4.

First, for arbitrary \(U\in\mathcal{T}_\fin\) and arbitrary \(i\), 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 arbitrary \(U\in\mathcal{T}\), each \(f^{-1}_i(U)\) must be open in \(Y_i\). Therefore, by the definition of \(\mathcal{T}_\fin\), we have \(U\in\mathcal{T}_\fin\), and thus \(\mathcal{T}_\fin\) is stronger than \(\mathcal{T}\).

Likewise, the final topology satisfies the following universal property, similar to that of the initial topology.

Proposition 6 In the situation of Definition 4, suppose additionally 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 opposite direction.

Suppose each function \(g\circ f_i\) is continuous. Then for any open set \(U\subseteq Z\), the 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.

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References

[Bou] N. Bourbaki, General Topology. Elements of mathematics. Springer, 1995.

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