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    The coarsest topology making a family of maps continuous

    E-UAMGV

    Area
    Topology
    Topics
    Product Topology, Topological Spaces
    Status
    draft

    Let \(A\) be a set; let \(\ts{X_\alpha}_{\alpha \in J}\) be an indexed family of spaces; and let \(\ts{f_\alpha}_{\alpha \in J}\) be an indexed family of functions \(f_\alpha: A \to X_\alpha\).

    1. Show there is a unique coarsest topology \(\mathcal{T}\) on \(A\) relative to which each of the functions \(f_\alpha\) is continuous.

    2. Let

    \[ \mathcal{S}_\beta = \ts{f_\beta^{-1}(U_\beta) \mid U_\beta \text{ is open in } X_\beta}, \]

    and let \(S = \bigcup \mathcal{S}_\beta\). Show that \(S\) is a subbasis for \(\mathcal{T}\).

    1. Show that a map \(g: Y \to A\) is continuous relative to \(\mathcal{T}\) if and only if each map \(f_\alpha \circ g\) is continuous.

    2. Let \(f: A \to \prod X_\alpha\) be defined by the equation

    \[ f(a) = (f_\alpha(a))_{\alpha \in J}; \]

    let \(Z\) denote the subspace \(f(A)\) of the product space \(\prod X_\alpha\). Show that the image under \(f\) of each element of \(\mathcal{T}\) is an open set of \(Z\).

    Solution

    1. Let \[ \mathcal S=\bigcup_{\beta\in J}\{f_\beta^{-1}(U):U\subseteq X_\beta\text{ open}\}. \] Let \(\mathcal T\) be the topology generated by the subbasis \(\mathcal S\). Then every \(f_\beta\) is continuous by construction. If \(\mathcal U\) is any topology on \(A\) making every \(f_\beta\) continuous, every member of \(\mathcal S\) belongs to \(\mathcal U\), hence the topology generated by \(\mathcal S\) satisfies \[ \mathcal T\subseteq\mathcal U. \] Therefore \(\mathcal T\) is the unique coarsest such topology.

    2. This is exactly the construction in part (a): \(\mathcal S\) is a subbasis for \(\mathcal T\).

    3. If \(g:Y\to A\) is continuous, then each \(f_\alpha\circ g\) is continuous as a composition. Conversely, suppose every \(f_\alpha\circ g\) is continuous. For a subbasic set \(f_\alpha^{-1}(U)\) of \(A\), \[ g^{-1}(f_\alpha^{-1}(U))=(f_\alpha\circ g)^{-1}(U), \] which is open in \(Y\). Therefore \(g\) is continuous.

    4. Define \[ f:A\to\prod_\alpha X_\alpha,\qquad f(a)=(f_\alpha(a))_\alpha, \] and let \(Z=f(A)\). The product topology on \(\prod X_\alpha\) has subbasis \(\pi_\alpha^{-1}(U)\), and \[ f^{-1}(\pi_\alpha^{-1}(U))=f_\alpha^{-1}(U). \] Thus \(\mathcal T\) is exactly the topology generated by inverse images under \(f\) of open sets of the product.

    Every \(\mathcal T\)-open set is therefore of the form \[ V=f^{-1}(W) \] for some product-open \(W\): this follows first for finite intersections of subbasic sets and then for arbitrary unions. Such \(V\) is saturated on fibers of \(f\), and \[ f(V)=f(A)\cap W=Z\cap W. \] Hence \(f(V)\) is open in the subspace \(Z\).

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