(a) Let $\mathfrak{T}$ be the collection of topologies on $A$ such that relative to each one each $f_{\alpha}$ is continuous. $\mathfrak{T}$ is nonempty, as there exists at least one topology on $A$ such that every function having $A$ as its domain is continuous, namely, the discrete topology. The intersection of an arbitrary subcollection of $\mathfrak{T}$ is a topology (Exercise 4(a) of §13) contained in $\mathfrak{T}$ (for every open set $U\subset X_{\alpha}$ , $f_{\alpha}^{-1}(U)$ is open in every topology in the subcollection, hence, it is open in their intersection). Therefore, $\mathcal{T}=\cap_{\mathcal{U}\in\mathfrak{T}}\mathcal{U}$ is the coarsest topology such that relative to it each of the functions $f_{\alpha}$ is continuous. The uniqueness follows immediately, as the constructed topology must be a subset of any other topology in $\mathfrak{T}$ .

(b) For every topology $\mathcal{U}\in\mathfrak{T}$ , every set in $\mathcal{S}$ must be open in $\mathcal{U}$ , as every $f_{\beta}$ is continuous relative to $\mathcal{U}$ . So, every topology in $\mathfrak{T}$ contains $\mathcal{S}$ . On the other hand, if a topology on $A$ contains $\mathcal{S}$ , then each $f_{\beta}$ is continuous relative to this topology. So, $\mathfrak{T}$ is exactly the collection of all topologies on $A$ containing $\mathcal{S}$ . The coarsest such topology $\mathcal{T}$ is generated by $\mathcal{S}$ as a subbasis (Exercise 5 of §13).

(c) Let $h_{\alpha}=f_{\alpha}\circ g$ . Then if $g$ is continuous, every $h_{\alpha}$ is continuous. If each $h_{\alpha}$ is continuous, then for every $U$ open in $A$ , $U$ is the union of finite intersections of some subbasis elements $f_{\beta}^{-1}(U_{\beta})$ , where each $U_{\beta}$ is open in $X_{\beta}$ . Therefore, $g^{-1}(U)$ is the union of finite intersections of $g^{-1}(f_{\beta}^{-1}(U_{\beta}))=h_{\beta}^{-1}(U_{\beta})$ , and, hence, open in $Y$ .

(d) Let $U\in\mathcal{T}$ and $\mathbf{x}\in f(U)$ . Take $a\in U$ such that $\mathbf{x}=f(a)$ . Since $U$ is open there is a basis element $V$ in $\mathcal{T}$ such that $a\in V\subset U$ and $V=\cap_{i=1,...,n}f_{\alpha_{i}}^{-1}(U_{\alpha_{i}})$ where $U_{\alpha_{i}}$ is open in $X_{\alpha_{i}}$ (without loss of generality all $\alpha$ ’s are assumed to be distinct). For all other $\alpha$ ’s assume $U_{\alpha}=X_{\alpha}$ . Then, $V=f^{-1}(\sqcap_{\alpha}U_{\alpha})$ and $f(V)=f(A)\cap\sqcap_{\alpha}U_{\alpha}$ . This set is open in the subspace topology (as only finitely many of $U$ ’s are distinct from their spaces) and $\mathbf{x}\in f(V)\subset f(U)$ . Therefore, $f(U)$ is open in the subspace topology.