Exercise 19.10

Proof. Let $C$ be the collection of topologies on $A$ relative to which each of the functions $f_{\alpha }$ is continuous. Clearly $C$ is nonempty as the discrete topology is in $C$ since every subset of $A$ is open in it so that $f_{\alpha }(V_{\alpha })$ is always open when $V_{\alpha }$ is open in $X_{\alpha }$. Let $T=\bigcap <!--\; FUNCTION\; APPLICATION\; -->C$, which is a topology on $A$ by what was shown in Exercise 13.4 part (a). We claim that this is the unique coarsest topology such that each $f_{\alpha }$ is continuous relative to it. To see this suppose that $T^{\prime }$ is any topology in such that each $f_{\alpha }$ is continuous relative to it, hence $T^{\prime }\in C$. Then, for any open $U\in T=\bigcap <!--\; FUNCTION\; APPLICATION\; -->C$ we of course have that $U\in T^{\prime }$ since $T^{\prime }\in C$. Hence $T\subset T^{\prime }$ since $U$ was arbitrary so that $T$ is courser than $T^{\prime }$, noting that it could of course be that $T=T^{\prime }$ as well. Since $T^{\prime }$ was artbitary, this shows the desired result.

Of course it also must be that $T$ is unique since, for any other $T^{\prime }$ that is a coarsest element of $C$, we just showed above that $T\subset T^{\prime }$ since $T^{\prime }\in C$. But also $T\supset T^{\prime }$ since $T^{\prime }$ must be coarser than $T$ since $T\in C$. This shows that $T=T^{\prime }$ so that $T$ is unique since $T^{\prime }$ was arbitrary. This also follows from the more general fact that any smallest element in an order or partial order is always unique, and inclusion is always at least a partial order. □

Proof. We show that $C$ from part (a) is exactly the set of topologies on $A$ that contain the subbasis $S$. That is, we show that $T^{\prime }\in C$ if and only if $S\subset T^{\prime }$ when $T^{\prime }$ is a topology on $A$. Since the coarsest topology $T$ from part (a) is defined as $\bigcap <!--\; FUNCTION\; APPLICATION\; -->C$, this shows that $T$ is the topology generated from the subbasis $S$ by Exercise 13.5.

$\text{(⇒)}$ Suppose that $T^{\prime }\in C$ so that every $f_{\alpha }$ is continuous relative to $T^{\prime }$. Now consider any subbasis element $S\in S$ so that $S={f_{\beta }}^{-1}(U_{\beta })$ for some $\beta \in J$ and some open set $U_{\beta }$ in $X_{\beta }$. Then $f_{\beta }$ is continuous relative to $T^{\prime }$ so that $S$ is open with respect to $T^{\prime }$, and hence $S\in T^{\prime }$. This shows that $S\subset T^{\prime }$ since $S$ was arbitrary, hence $T^{\prime }$ contains $S$.

$\text{(⇐)}$ Now suppose that $T^{\prime }$ is a topology on $A$ that contains $S$ so that $S\subset T^{\prime }$. Consider any $\alpha \in J$ and any open set $U_{\alpha }$ of $X_{\alpha }$. Then clearly ${f_{\alpha }}^{-1}(U_{\alpha })$ is in $S_{\alpha }$ so that it is also clearly in $S=\bigcup <!--\; FUNCTION\; APPLICATION\; -->S_{\beta }$. Hence also ${f_{\alpha }}^{-1}(U_{\alpha })\in T^{\prime }$ since $S\subset T^{\prime }$. Therefore ${f_{\alpha }}^{-1}(U_{\alpha })$ is open with respect to $T^{\prime }$, which shows that $f_{\alpha }$ is continuous relative to $T^{\prime }$ since $U_{\alpha }$ was an arbitrary open set of $X_{\alpha }$. Since $\alpha \in J$ was also arbitrary, this shows that every $f_{\alpha }$ is continuous relative to $T^{\prime }$ so that $T^{\prime }\in C$ by definition. □

Proof. $\text{(⇒)}$ Suppose that $g:Y\to A$ is continuous relative to $T$. Consider any $\alpha \in J$ and any open set $U_{\alpha }$ of $X_{\alpha }$. Then ${f_{\alpha }}^{-1}(U_{\alpha })$ is open with respect to $T$ since $f_{\alpha }$ is continuous relative to $T$ since every $f_{\alpha }$ is. It then follows that $g^{-1}({f_{\alpha }}^{-1}(U_{\alpha }))$ is open in $Y$ since $g$ is continuous relative to $T$. From Exercise 2.4 part (a) we have that $g^{-1}({f_{\alpha }}^{-1}(U_{\alpha }))={(f_{\alpha }\circ g)}^{-1}(U_{\alpha })$, which shows that $f_{\alpha }\circ g$ is continuous since $U_{\alpha }$ was an arbitrary open set of $X_{\alpha }$. Since $\alpha \in J$ was arbitrary, this shows the desired result.

$\text{(⇐)}$ Now suppose that every $f_{\alpha }\circ g$ is continuous and consider any open set $U$ of $A$ with respect to $T$. Then by part (b) we have that $U$ is an arbitrary union of finite intersections of subbasis elements ${f_{\alpha }}^{-1}(U_{\alpha })$ for $\alpha \in J$ and open $U_{\alpha }$ in $X_{\alpha }$. It then follows from Exercise 2.2 parts (b) and (c) that $g^{-1}(U)$ is an arbitrary union of finite intersections of sets $g^{-1}({f_{\alpha }}^{-1}(U_{\alpha }))$. Again we have that each $g^{-1}({f_{\alpha }}^{-1}(U_{\alpha }))={(f_{\alpha }\circ g)}^{-1}(U_{\alpha })$ by Exercise 2.4 part (a) so that each of these sets is open in $Y$ since every $f_{\alpha }\circ g$ is continuous. Hence $g^{-1}(U)$ is open as well since it is the arbitrary union of finite intersections of these open sets and $Y$ is a topological space. Since $U$ was an arbitrary open set of $A$ with respect to $T$, this shows that $g$ is continuous relative to $T$ as desired. □

Proof. Suppose that $U$ is any open set of $A$ with respect to $T$. Consider any $y={(y_{\alpha })}_{\alpha \in J}\in f(U)$ so that there is an $a\in U$ where $f(a)=y$. Since $a\in U$ and $U$ is open in $A$, we have that there is a basis element $B_A$ containing $a$ where $B_A\subset U$. It then follows from part (b) that this basis element is a finite intersection of subbasis elements, hence $B_A={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{\beta \in I}{f_{\beta }}^{-1}(U_{\beta })$, where $I\subset J$ is finite and each $U_{\beta }$ is open in $X_{\beta }$. Now define

so that clearly the set $B_p=\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$ is a basis element of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$ in the product topology by Theorem 19.1 since $I$ is finite. We then have that $B_Z=Z\cap B_p$ is a basis element of the subspace $Z$ by Lemma 16.1.

Now, we have that $a\in U$ and $U\subset A$ so that $a\in A$ as well. It then follows that $y=f(a)\in f(A)=Z$. For $\beta \in I$, we also have that $a\in {f_{\beta }}^{-1}(U_{\beta })$ since the basis element $B_A={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{\beta \in I}{f_{\beta }}^{-1}(U_{\beta })$ contains $a$. Hence $f_{\beta }(a)\in U_{\beta }$. Since of course every other $f_{\alpha }(a)\in X_{\alpha }$ when $\alpha \notin I$, we have that $f_{\alpha }(a)\in V_{\alpha }$ for all $\alpha \in J$ and thus $y=f(a)={(f_{\alpha }(a))}_{\alpha \in J}\in \prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }=B_p$. We therefore have that $y\in Z\cap B_p=B_Z$ so that $B_Z$ contains $y$.

Lastly, consider any $z={(z_{\alpha })}_{\alpha \in J}\in B_Z=Z\cap B_p$. Then $z\in Z=f(A)$ so that there is an $x\in A$ where $f(x)={(f_{\alpha }(x))}_{\alpha \in J}=z$ and hence each $f_{\alpha }(x)=z_{\alpha }$. We also have that $z\in B_p=\prod <!--\; FUNCTION\; APPLICATION\; -->V_{\alpha }$ so that $z_{\alpha }\in V_{\alpha }$ for every $\alpha \in J$. In particular $f_{\beta }(x)=z_{\beta }\in V_{\beta }=U_{\beta }$ for all $\beta \in I$ so that $x\in {f_{\beta }}^{-1}(U_{\beta })$. Therefore $x\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{\beta \in I}{f_{\beta }}^{-1}(U_{\beta })=B_A$ so that also $x\in U$ since $B_A\subset U$. Then we have that $z=f(x)\in f(U)$. Since $z$ was arbitrary this shows that $B_Z\subset f(U)$.

We have thus shown that $B_Z$ is a basis element of the subspace $Z$ that contains $y$ where $B_Z\subset f(U)$. Since $y$ was an arbitrary element of $f(U)$, this suffices to show that $f(U)$ is open in the subspace $Z$ as desired. □