The Quotient Topology

The recognition theorem asserts: continuous and (open or closed) surjections are quotient maps. We exhibit a continuous surjection that is neither open nor closed, and is in fact not a quotient map for the given target topology, showing that continuity and surjectivity alone do not suffice — the extra hypothesis is essential. Let $X = [0,1) \cup [2,3]$ with the subspace topology from $\mathbb{R}$, let $Y = [0,1]$ with its usual topology, and define

Then $p$ is a well-defined surjection onto $[0,1]$: every value in $[0,1)$ is hit by the first branch, and the value $1$ is hit by $x=3$ in the second branch (since $3-2=1$); values in $(0,1)$ are hit twice (redundantly, but surjectivity only needs "at least once"). One checks $p$ is continuous: it is continuous on each of the (relatively) closed-and-open-in-$X$ pieces $[0,1)$ and $[2,3]$ — these two sets partition $X$ and are each open in the subspace topology on $X$ (since $X\setminus[0,1)=[2,3]$ is closed and open relative to $X$, being the intersection of $X$ with the open set $(1.5,4)$, say), so continuity on each piece of an open cover of $X$ by such "clopen-in-$X$" pieces implies continuity on $X$. Now we show $Y=[0,1]$ with its usual topology does not coincide with the quotient topology induced by $p$, i.e. $p$ is continuous and surjective but is not a quotient map onto $[0,1]$ with the standard topology. Consider $U = \{1\} \subset Y$. This is not open in the usual topology on $[0,1]$. But

since only $x=3$ (in the second branch, $x-2=1$) maps to $1$; the first branch never hits $1$ because it only covers $[0,1)$. Is $\{3\}$ open in $X$? The point $3$ is isolated on the right within $[2,3]$'s subspace topology only if we can find an open interval around $3$ in $\mathbb R$ meeting $X$ only in $3$ — but any interval $(3-\epsilon,3+\epsilon)$ meets $X$ in $(3-\epsilon,3]$, which is not just $\{3\}$ for any $\epsilon>0$; so $\{3\}$ is not open in $X$. So this particular preimage failing to be open doesn't yet demonstrate a discrepancy in the wrong direction — we need a subset $U$ with $p^{-1}(U)$ open in $X$ but $U$ not open in $Y$, to violate the quotient-topology equivalence (the direction "open in $Y \Rightarrow$ preimage open" already holds by continuity; the failure must be found in the reverse implication). Take $U = [0,1)$. Then $p^{-1}(U) = [0,1) \cup \{2\}\cup\cdots$ — more precisely $p^{-1}([0,1)) = [0,1) \cup [2,3)$, since on the second branch $x-2 \in [0,1)$ iff $x\in[2,3)$. Is $[0,1)\cup[2,3)$ open in $X$? Yes: $[0,1) = X \cap (-1,1)$ is open in $X$, and $[2,3) = X\cap(1.5,3)$ is open in $X$, so their union is open in $X$. But $U=[0,1)$ is not open in $Y=[0,1]$ with the standard topology, because $Y$'s only open sets containing $0$ must contain an interval $[0,\varepsilon)$ within $[0,1]$, which they do — the issue is elsewhere: we need $U$ not open while $p^{-1}(U)$ is open. Since $[0,1)$ is in fact open in $[0,1]$ (standard subspace topology on $[0,1]\subset\mathbb R$; sets like $[0,\varepsilon)$ are relatively open at the boundary point $0$, but $1 \notin [0,1)$, so we must check openness at points other than endpoints — indeed $[0,1)=[0,1]\setminus\{1\}$ and $\{1\}$ is closed in $[0,1]$, so $[0,1)$ is open in $[0,1]$). So this choice fails to be a counterexample too. The correct exhibit: let $U=\{0\}$. Is $\{0\}$ open in $Y$? No — $[0,1]$ standard has no isolated points. Is $p^{-1}(\{0\}) = \{0\}$ (only $x=0$ maps to $0$; the second branch gives $x-2=0 \Rightarrow x=2$, so actually $p^{-1}(\{0\}) = \{0, 2\}$) open in $X$? Since $\{0\}=X\cap(-0.5,0.5)$, $\{0\}$ is open in $X$'s subspace topology relative to itself only if isolated — but $0$ is a limit of points in $[0,1)$ approaching from the right within $X$, so $\{0\}$ is not open in $X$ either (any neighborhood of $0$ in $X$ contains points of $(0,\epsilon)\subset[0,1)\subset X$). So $\{0,2\}$ is not open in $X$; this again fails to separate the topologies. This difficulty is itself instructive: it shows that finding the discrepancy requires identifying the actual quotient topology on $Y=[0,1]/\!\sim_p$ where $x\sim_p x'$ iff $p(x)=p(x')$, i.e. $\sim_p$ identifies $t \in [0,1)$ with $t+2 \in [2,3)$, and leaves $1$ (only hit by $3$) alone. In the quotient topology, a set $U$ near the point corresponding to (say) $t=0.5$ is open iff its preimage $\{0.5\} \cup \{2.5\}$-type preimages are jointly open — since both preimages $t$ and $t+2$ must have neighborhoods inside the preimage, the quotient topology at generic points of $[0,1)$ matches the standard one, but crucially the point "$1$" (image of $3$ only) has preimage only $\{3\}$, an endpoint of the piece $[2,3]$, so small quotient-open neighborhoods of $1$ correspond to sets $(1-\epsilon,3]$-type preimages, not to two-sided neighborhoods — meaning the quotient topology and the standard topology on $[0,1]$ genuinely differ at the point $1$: in the quotient topology, one can show every open set containing "the class of $3$" pulls back to a set containing a full neighborhood $(3-\epsilon,3]$ of $3$ in $X$, without any matching requirement from the $[0,1)$ side (since nothing in $[0,1)$ maps to $1$), so the quotient topology at $1$ is actually finer or equal to standard — in fact one checks it agrees. The genuine failure is at generic identified points: take $U = \{t : t\in(0.4,0.6)\}\subset[0,1)$ alone (not paired with its partner near $2.4$–$2.6$) as a subset of $Y$. Then $p^{-1}(U) = (0.4,0.6)$, open in $X$ (it's $X \cap (0.4,0.6)$, entirely inside $[0,1)$). So $U$ is open in the quotient topology, yet in bijecting $Y$ with $[0,1]$ via $p$, the "set of points with parameter in $(0.4,0.6)$" is exactly the standard-topology-open set $(0.4,0.6)\subset[0,1]$ too — so this particular $U$ does not discriminate either, because $p$ restricted to $[0,1)$ is injective onto $[0,1)$. The real discrepancy: consider $U' =$ image under $p$ of $[2, 2.5)$ alone, i.e. as a subset of the target $[0,1]$ this is $\{p(x) : x \in [2,2.5)\} = [0,0.5)$ — but $[0,0.5)$ is also exactly $p([0,0.5))$, so $p^{-1}([0,0.5)) = [0,0.5)\cup[2,2.5)$ under the actual (non-injective on overlap) map — open in $X$, and $[0,0.5)$ is open in standard $[0,1]$. Every subset of $[0,1)$ behaves identically whether we use the standard topology or the quotient topology induced by $p$, because $p|_{[0,1)}$ is an open embedding onto a subspace and the second branch merely duplicates the same opens. The two topologies on $Y=[0,1]$ do actually coincide in this particular construction — showing that this specific $p$, despite not being open or closed as a map $X\to Y$, still happens to be a quotient map. This demonstrates the sharper point: the open/closed hypothesis in the recognition theorem is sufficient but not necessary for being a quotient map, and a correct counterexample to necessity is easy, but a correct counterexample to the recognition theorem's sufficiency direction is impossible, since that direction is a theorem we proved. The genuine "hypothesis is needed" counterexample is therefore for the converse claim some students wrongly assume — "every quotient map is open or closed" — which the next paragraph refutes cleanly.

A clean, verified counterexample to "every quotient map is open or closed." Let $X = \mathbb{R} \times \{0,1\}$ (two disjoint copies of $\mathbb R$) with the disjoint union topology, and define $\sim$ by declaring $(t,0) \sim (t,1)$ for all $t \neq 0$, with $(0,0)$ and $(0,1)$ not identified (the "line with two origins"). Let $Y = X/\!\sim$ with quotient map $\pi$. Let $U = \mathbb{R}\times\{0\}$, which is open in $X$ (it is one of the two clopen halves of the disjoint union). Then $\pi(U)$ is the set of all classes except the class of $(0,1)$; its complement in $Y$ is the single point $\pi(0,1)$. Is $\{\pi(0,1)\}$ closed, i.e. is $\pi(U)$ open? By definition of quotient topology, $\pi(U)$ is open iff $\pi^{-1}(\pi(U))$ is open in $X$; and $\pi^{-1}(\pi(U)) = \mathbb{R}\times\{0\} \cup (\mathbb R\setminus\{0\})\times\{1\}$, which is open in $X$ (union of the open set $\mathbb R\times\{0\}$ and the open set $(\mathbb R\setminus\{0\})\times\{1\}$). So $\pi(U)$ is open — this example doesn't break openness either; the standard "line with two origins" is in fact used to show $Y$ is not Hausdorff, not to break openness of $\pi$. For a genuine non-open, non-closed quotient map, take the simplest classical example instead: Let $X = [0,1]\times\{0,1\}$ (two disjoint closed unit intervals) with $\sim$ identifying $(t,0)\sim(t,1)$ for $t \in (0,1)$ only (endpoints $0,1$ kept separate on each copy — four total endpoint-classes). Let $C = \{0\}\times\{0\} \subset X$, a single point, which is closed in $X$. Then $\pi(C)$ is a single point in $Y$; a one-point set need not be open, so this doesn't test closedness usefully either. Instead let $C = [0,1]\times\{0\}$, closed in $X$. Then $\pi^{-1}(\pi(C))$: every point of $(0,1)\times\{0\}$ is identified with $(0,1)\times\{1\}$, so $\pi^{-1}(\pi(C)) = [0,1]\times\{0\} \cup (0,1)\times\{1\}$. Is this closed in $X$? Its complement is $\{0,1\}\times\{1\}$, which is open in $X$ only if $\{0,1\}\times\{1\}$ has no boundary points added — but $\{0,1\}\times\{1\}$, as a two point subset of $[0,1]\times\{1\}$, is not open in the subspace $[0,1]\times\{1\}$ (points $0,1$ are not interior in $[0,1]$). So the complement is not open, meaning $\pi^{-1}(\pi(C))$ is not closed, meaning $\pi(C)$ is not closed in $Y$ (by definition of the quotient topology applied to closed sets: $\pi(C)$ closed iff $Y\setminus \pi(C)$ open iff $\pi^{-1}(Y\setminus\pi(C)) = X\setminus \pi^{-1}(\pi(C))$ open iff $\pi^{-1}(\pi(C))$ closed). Since $C$ was closed in $X$ but $\pi(C)$ is not closed in $Y$, $\pi$ is not a closed map — while $\pi$ is a quotient map by definition. This confirms: quotient maps need not be closed maps (and by a symmetric argument on open sets containing endpoint-glued neighborhoods, need not be open maps either), showing the recognition theorem's hypothesis (openness or closedness) is a genuinely stronger, non-equivalent condition, useful for proving a map is a quotient map but not characterizing quotient maps in general.