* (a) $Z$ is not locally compact. Example 7 of §22. Let $p$ be a quotient map on $X=\mathbb{R}$ that identifies all positive integers to a point $b$ . Let $Z=\mathbb{Q}$ (it is not locally compact but Hausdorff). Then $\pi=p\times\iota_Z$ is not a quotient map. The open set $U$ in the example has the property that for any $\delta>0$ and $\epsilon>0$ there is a large enough $n\in p^{-1}(b)$ such that $(n+\epsilon,\delta)$ is not in $U$ . In other words, there is no saturated neighborhood $U_1$ of, let say, point $1$ in $\mathbb{R}$ for which there exists neighborhood $V$ of $0$ in $\mathbb{Q}$ such that there product $U_1\times V$ is contained in $U$ . Any such neighborhood $U_1$ being saturated would contain all points $n$ and some their local basis neighborhoods $W_n=(n-\epsilon_n,n+\epsilon_n)$ . Any $V$ would contain an interval $(-\delta,+\delta)\cap\mathbb{Q}$ . Then for a large enough $n$ such that $c_n<\delta$ there would be a point in $(n-\epsilon_n,n+\epsilon_n)\times(-\delta,+\delta)\cap\mathbb{Q}$ not in $U$ . Note that $U$ is a neighborhood such that it contains $n\times\mathbb{Q}$ for any $n$ but does not contain any tube about it. This would not be possible if $\mathbb{Q}$ were compact.$Z$ is not Hausdorff. Exercise 6 of §22 is an example of a quotient map from a non-locally compact Hausdorff set onto a compact non-Hausdorff space such that the product of it with itself is not a quotient map. What I want is similar to the previous example with the only difference that $Z$ is locally compact but not Hausdorff. Let us take a one point compactification of $\mathbb{Q}$ (in fact, the one point compactification was defined for locally compact Hausdorff spaces only, $\mathbb{Q}$ is not locally compact, but it seems that the Hausdorff property is enough to construct the one-point compactification, and local compactness is needed for the Hausdorff property of the constructed space only). Therefore, the constructed space $\mathbb{Q}'$ is (locally) compact but not Hausdorff: the open sets containing the new point are those having bounded and closed in $\mathbb{R}$ complement, such sets have no interior points in $\mathbb{Q}$ . Now, if we take the same example, then $U$ is still saturated and open, and its image is not open. The difference between this case and the previous is that here we made the space compact at the cost of being Hausdorff, hence, we still cannot guarantee that any neighborhood would contain a compact closure of another neighborhood — something that we need for the proof.Now the proof. $\pi$ is continuous (see §18). Let $B=p^{-1}(A)$ be open. We show that $A$ is open. Let $(x,z)=p^{-1}((y,z))$ . Since $B$ is open and $Z$ is locally compact there are neighborhoods $U_1$ of $x$ in $X$ and $V$ of $z$ in $Z$ such that $U_1\times\overline{V}\subseteq B$ . If we knew that $U_1$ is saturated (or, alternatively, if we knew that $p$ is an open map) we would immediately conclude that $(y,z)$ has a neighborhood contained in $A$ . Since we don’t know that, we must proceed in finding a saturated neighborhood $U$ of $x$ such that $U\times V\subseteq B$ . Since $B$ is saturated, we have $U_1\times\overline{V}\subseteq p^{-1}(p(U_1))\times\overline{V}\subseteq B$ . In the first example, suppose we take point (1,0), then find a "rectangle neighborhood" about it, then the image of this rectangle has point $b$ , therefore, the preimage of the image has all points $(n,0)$ (the preimage of the image of the rectangle neighborhood contains the original rectangle about (1,0) plus all other points $(n,0)$ with vertical intervals). Now we use the fact that $\overline{V}$ is compact. For each point in $p^{-1}(p(U_1))$ we take a neighborhood such that its product with $\overline{V}$ is still in $B$ . We call the union of all this neighborhoods $U_2$ . This is exactly what we could not do in the first example: for any $\overline{V}$ there was a large enough point $n$ such that we could not find the desired neighborhood containing it due to the fact that no $\overline{V}$ were compact. Now we proceed this way constructing $U_3, U_4, . . .$ . Note that for each $n$ , $U_n\subseteq p^{-1}(p(U_n))\subseteq U_{n+1}$ and $U_n\times\overline{V}\subseteq B$ . Let $U=\cup_n U_n$ . Then $U\times V\subseteq B$ is open. If we show that $U$ is saturated, we are done. Indeed, $U\subseteq$ $p^{-1}(p(U))=$ $\cup_n p^{-1}(p(U_n))\subseteq$ $\cup_n U_{n+1}=U$ , hence $U=p^{-1}(p(U))$ is saturated.(b) This is the easier part. Using (a), we conclude that the composition of two quotient maps $p\times q=(\iota_B\times q)\circ(p\times\iota_C)$ is a quotient map given that both $B$ and $C$ are locally compact and Hausdorff.