[proofplan] We prove both assertions. First, we show the product [topology](/page/Topology) is the coarsest topology making all projections continuous by identifying the minimal collection of [open sets](/page/Open%20Set) forced by [continuity](/page/Continuity) of each $\pi_\beta$ (the preimages $\pi_\beta^{-1}(V_\beta)$ for $V_\beta \in \tau_\beta$), observing that these form a subbasis whose generated topology is exactly the product topology. Second, we prove the universal property: $f: Z \to X$ is continuous if and only if each $\pi_\alpha \circ f$ is continuous. The forward direction follows from composition of continuous maps. The reverse direction uses the subbasis characterisation: preimages of subbasis elements under $f$ are open precisely when each $\pi_\alpha \circ f$ is continuous. [/proofplan] [step:Identify the open sets forced by continuity of the projections] For each $\beta \in A$, the projection map is \begin{align*} \pi_\beta: \prod_{\alpha \in A} X_\alpha &\to X_\beta \\ (x_\alpha)_{\alpha \in A} &\mapsto x_\beta. \end{align*} If $\tau$ is any [topology](/page/Topology) on $X = \prod_{\alpha \in A} X_\alpha$ making $\pi_\beta$ [continuous](/page/Continuity), then $\pi_\beta^{-1}(V_\beta) \in \tau$ for every $V_\beta \in \tau_\beta$. Consequently, $\tau$ must contain the collection \begin{align*} \mathcal{S} = \bigl\{\pi_\beta^{-1}(V_\beta) : \beta \in A,\, V_\beta \in \tau_\beta\bigr\}. \end{align*} Since $\tau$ is a topology, it must also contain all finite intersections of elements of $\mathcal{S}$ and all unions of such finite intersections. In other words, $\tau$ must contain the topology generated by $\mathcal{S}$ as a subbasis. [guided] The question driving this step is: what is the *least* we must include in a topology on $X$ if we insist that every projection $\pi_\beta$ is continuous? Continuity of $\pi_\beta$ means precisely that $\pi_\beta^{-1}(V_\beta)$ is open in $X$ for every open $V_\beta \subset X_\beta$. So the collection $\mathcal{S}$ of all such preimages must be contained in any topology making all projections continuous. The topology generated by $\mathcal{S}$ — i.e., the smallest topology containing $\mathcal{S}$ — is therefore the coarsest candidate. Explicitly, a set $\pi_\beta^{-1}(V_\beta)$ consists of all tuples $(x_\alpha)_{\alpha \in A}$ whose $\beta$-th coordinate lies in $V_\beta$, with no restriction on any other coordinate. These are the "slabs" in the product that constrain a single coordinate. [/guided] [/step] [step:Verify that $\mathcal{S}$ generates exactly the product topology] The product [topology](/page/Topology) $\tau_{\mathrm{prod}}$ is defined to have the basis \begin{align*} \mathcal{B} = \Bigl\{\prod_{\alpha \in A} U_\alpha : U_\alpha \in \tau_\alpha \text{ for all } \alpha,\; U_\alpha = X_\alpha \text{ for all but finitely many } \alpha\Bigr\}. \end{align*} Each basis element $\prod_{\alpha \in A} U_\alpha \in \mathcal{B}$ can be written as a finite intersection of subbasis elements: if $\alpha_1, \ldots, \alpha_n$ are the indices where $U_{\alpha_i} \neq X_{\alpha_i}$, then \begin{align*} \prod_{\alpha \in A} U_\alpha = \pi_{\alpha_1}^{-1}(U_{\alpha_1}) \cap \cdots \cap \pi_{\alpha_n}^{-1}(U_{\alpha_n}). \end{align*} Conversely, each subbasis element $\pi_\beta^{-1}(V_\beta)$ is itself a basis element (take $U_\alpha = X_\alpha$ for $\alpha \neq \beta$ and $U_\beta = V_\beta$). Therefore $\mathcal{S} \subset \mathcal{B}$, and every element of $\mathcal{B}$ is a finite intersection of elements of $\mathcal{S}$. It follows that $\mathcal{S}$ is a subbasis for $\tau_{\mathrm{prod}}$, i.e., the topology generated by $\mathcal{S}$ is exactly $\tau_{\mathrm{prod}}$. [guided] We need to confirm that the subbasis $\mathcal{S}$ generates precisely the product topology, not something larger or smaller. A basis element in $\mathcal{B}$ is a "box" $\prod_{\alpha} U_\alpha$ where all but finitely many factors are the full space $X_\alpha$. If the constrained indices are $\alpha_1, \ldots, \alpha_n$, then \begin{align*} \prod_{\alpha \in A} U_\alpha &= \{(x_\alpha)_{\alpha \in A} : x_{\alpha_i} \in U_{\alpha_i} \text{ for } i = 1, \ldots, n\} \\ &= \bigcap_{i=1}^n \{(x_\alpha)_{\alpha \in A} : x_{\alpha_i} \in U_{\alpha_i}\} \\ &= \bigcap_{i=1}^n \pi_{\alpha_i}^{-1}(U_{\alpha_i}). \end{align*} So every basis element is a finite intersection of subbasis elements. Conversely, each $\pi_\beta^{-1}(V_\beta)$ corresponds to the box with $U_\beta = V_\beta$ and $U_\alpha = X_\alpha$ for $\alpha \neq \beta$, which is itself a basis element. Therefore $\mathcal{S}$ is a subbasis for $\tau_{\mathrm{prod}}$, and the topology generated by $\mathcal{S}$ coincides with the product topology. [/guided] [/step] [step:Conclude that the product topology is the coarsest topology making all projections continuous] By the first step, any [topology](/page/Topology) making all projections [continuous](/page/Continuity) must contain $\mathcal{S}$ and hence must contain the topology generated by $\mathcal{S}$. By the second step, this generated topology is $\tau_{\mathrm{prod}}$. Therefore $\tau_{\mathrm{prod}} \subset \tau$ for every such topology $\tau$. Moreover, $\tau_{\mathrm{prod}}$ itself makes all projections continuous: for each $\beta \in A$ and $V_\beta \in \tau_\beta$, the preimage $\pi_\beta^{-1}(V_\beta) \in \mathcal{S} \subset \tau_{\mathrm{prod}}$. So $\tau_{\mathrm{prod}}$ is the coarsest topology with this property. [/step] [step:Prove the forward direction of the universal property: continuity of $f$ implies continuity of each $\pi_\alpha \circ f$] Let $Z$ be a [topological](/page/Topology) space and let $f: Z \to X$ be [continuous](/page/Continuity). For each $\alpha \in A$, the projection $\pi_\alpha: X \to X_\alpha$ is continuous (as shown above). The composition of continuous maps is continuous, so $\pi_\alpha \circ f: Z \to X_\alpha$ is continuous. [/step] [step:Prove the reverse direction: continuity of each $\pi_\alpha \circ f$ implies continuity of $f$] Assume $\pi_\alpha \circ f: Z \to X_\alpha$ is [continuous](/page/Continuity) for every $\alpha \in A$. To show $f: Z \to X$ is continuous, it suffices to show that $f^{-1}(S) \subset Z$ is open for every $S$ in the subbasis $\mathcal{S}$ of the product [topology](/page/Topology). (A map is continuous if and only if the preimage of every subbasis element is open, since every [open set](/page/Open%20Set) is a union of finite intersections of subbasis elements, and preimages commute with unions and intersections.) Each subbasis element has the form $S = \pi_\beta^{-1}(V_\beta)$ for some $\beta \in A$ and $V_\beta \in \tau_\beta$. Compute: \begin{align*} f^{-1}\bigl(\pi_\beta^{-1}(V_\beta)\bigr) = (\pi_\beta \circ f)^{-1}(V_\beta). \end{align*} Since $\pi_\beta \circ f: Z \to X_\beta$ is continuous by hypothesis and $V_\beta \in \tau_\beta$ is open, the set $(\pi_\beta \circ f)^{-1}(V_\beta)$ is open in $Z$. Therefore $f^{-1}(S)$ is open in $Z$ for every $S \in \mathcal{S}$, so $f$ is continuous. [guided] This direction is where the universal property earns its name: to check continuity of a map *into* a product, it suffices to check continuity of each component. The proof reduces to one algebraic identity about preimages: \begin{align*} f^{-1}(\pi_\beta^{-1}(V_\beta)) = (\pi_\beta \circ f)^{-1}(V_\beta). \end{align*} This holds because both sides equal $\{z \in Z : \pi_\beta(f(z)) \in V_\beta\}$. The subbasis criterion for continuity — a map is continuous if and only if preimages of subbasis elements are open — is the key technical tool. Why does checking the subbasis suffice? Because every open set in the product topology is a union of finite intersections of subbasis elements. Preimages preserve both unions ($f^{-1}(\bigcup U_i) = \bigcup f^{-1}(U_i)$) and intersections ($f^{-1}(\bigcap U_i) = \bigcap f^{-1}(U_i)$), so if preimages of subbasis elements are open, preimages of all open sets are open. This characterisation makes the product topology the *categorical product* in the category of topological spaces: it is defined precisely so that maps into the product correspond bijectively to families of maps into the factors, with continuity matching on both sides. Any finer topology on $\prod_\alpha X_\alpha$ would make the projections continuous but would require "more" of a map $f: Z \to X$ for it to be continuous — potentially breaking the equivalence. [/guided] [/step]