I'm studying Topology for the first time, using Munkres(self study). Some time ago I first saw the definition of a continuous function:

\begin{equation} f:X\to Y \text{ is continuous iff for every open set }V\subset Y, f^{-1}(V) \text{ is open in X}. \end{equation}

I had no clue what this definition meant, until I saw this equivalent definition:

$f: X\to Y$ is continuous iff $\forall x\in X $ and for every open set $V$ containing $f(x)$, $\exists U\text{ s.t. }x\in U\text{ and }f(U) \subset V$, where $U$ is open.

which seems like a very natural generalization of continuity studied in analysis. So my first question is, why would you write the definition in the first form? Does this become natural after I study a lot of topology? It seems a little obscure to me.

Also, today I had a similar situation with the definition of the Quotient topology:

Definition. Let $X$ and $Y$ be topological spaces; let $p:X\to Y$ be a surjective map. The map $p$ is said to be a quotient map provided a subset $U$ of $Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$.

This time I read it, and I read the paragraph after the definition which said about saturated sets. But this time, I really have absolutely no clue about this definition that I can't read more. I would like to know what this definition has to do with the quotient I saw in group theory or such.

  • 6
    $\begingroup$ Isn't the first form much cleaner? $\endgroup$
    – Randall
    Aug 29, 2019 at 14:34
  • 2
    $\begingroup$ I think it's very common for an author to introduce a definition in its “cleanest” form first, then follow up with a remark on an equivalent definition that is less clean but more understandable. $\endgroup$ Aug 29, 2019 at 14:47
  • 2
    $\begingroup$ @Matthew Oh ok.. It just seemed unnatural at first. Better get used to it. $\endgroup$
    – Quasi07
    Aug 29, 2019 at 14:50
  • 1
    $\begingroup$ @Quasi07 in my solitary opinion, yes, that is the reason. I like it because it defines continuity globally without reference to particular points $x$. Everything is a matter of taste, of course. $\endgroup$
    – Randall
    Aug 29, 2019 at 14:50
  • 4
    $\begingroup$ To add another comment, the first form is also more suitable to arguments in point-set topology. Proving that connectedness is an invariant is much easier with the first version than the second, e.g. $\endgroup$
    – Randall
    Aug 29, 2019 at 14:55

2 Answers 2


So my first question is, why would you write the definition in the first form? Does this become natural after I study a lot of topology? It seems a little obscure to me.

It does become natural after a while, as it happens with most things that one works with for an extended period of time.

Regarding the definition itself, there is an excellent question on MathOverflow on a related topic.

One answer that I find satisfying says that, paraphrasing, all the concepts in general topology can be stated in terms of open sets, which then make defining a topology in terms of open sets an efficient approach. This is not to say one should always think of continuity, compactness, connectedness, etc. in terms of open sets.

I would like to know what this definition has to do with the quotient I saw in group theory or such.

I think the more intuitive way to think about this is to first define quotients of topological spaces. So let's give a motivation for that first, in analogy with their group theoretic counterpart.

Take a group $G$ and a normal subgroup $H$. The quotient group $G/H$ is defined as the equivalence classes of the relation $g \sim g' \iff gg^{-1} \in H$,

$$ G/H = \{gH : a \in G\} $$

toghether with the operation $gH \cdot g'H = gg'H$. Since $H$ is normal, this defines a group, and we have a canonical group morphism

$$ q : G \to G/H $$

that sends an element $g$ of $G$ to its class $gH$.

In general, given a group morphism $f : G \to G'$, its kernel is always a normal subgroup of $G$ and thus we can make sense of the quotient $G/\ker f$. Moreover, the first isomorphism theorem for groups says that

$$ G/\ker f \simeq \operatorname{im} f. $$


$$ \bar{f} : [g] \in G/\ker f \mapsto f(g) \in \operatorname{im} f \subset G' $$

Intuitively, the elements on the image are like the elements of $G$, up to multiplication by elements that are "trivial for $f$". When $f$ is surjective, we obtain

$$ G/\ker f \simeq G'. $$

Now, can we describe the equivalence relation previously defined in terms of $f$. Two elements $[g],[g']$ are equal in $G/\ker f$ precisely when $gg'^{-1} \in \ker f$, that is, when $1 = f(gg'^{-1}) = f(g)f(g')^{-1}$. In other terms,

$$ g \sim g' \iff f(g) = f(g'). $$

With this terminology, we know that for any group morphism $f : G \to G'$ we can define the equivalence relation $g \sim g' \iff f(g) = f(g')$, the equivalence classes have a group structure, and they are isomorphic to $\operatorname{im} f$.

Now let's go back to topological spaces. Take a space $X$ and an equivalence relation $R$ on $X$. We can define a topology on the equivalence classes $X/R$ by giving them the final topology with respect to the canonical projection $q : x \in X \mapsto [x] \in X/R$, that is, a set $U \subset X/R$ is open if and only if $q^{-1}(U)$ is open in $X$. One may argue that this is somewhat natural, as we want $q$ to be continuous and this topology is the minimal one that satisfies this.

Now, if we have a continuous $f : X \to Y$ that is compatible with $R$, i.e. such that $x \sim y$ implies $f(x) = f(y)$, we obtain a countinuous function

$$ \bar{f} : [x] \in X/R \mapsto f(x) \in Y $$

This is suspiciously similar to what we can do for groups. In effect, if $f : X \to Y$ is a continuous map, we can define an equivalence relation $R$ as

$$ x \sim y \iff f(x) = f(y) $$

exactly as we did we groups. Then $f$ is trivially compatible with $R$ and thus we obtain a map

$$ \bar{f} : [x] \in X/R \mapsto f(x) \in f(X) \subset Y, $$

again as in the group theoretic case. It would be nice to recover a version of the first isomorphism theorem in this setting, i.e. it would be nice for $\bar{f} : X \to f(X)$ to be a homeomorphism.

Since we are restricting ourselves to $f(X)$ anyway, we can assume $f$ surjective. It can be shown that $\bar{f}$ is bijective (the proof is identical to the one for groups), and we already know that $\bar{f}$ is continuous, so what is left to determine is whether $\bar{f}$ is open.

Take an open set $U \subset X/R$. By definition $U$ is open if and only if $q^{-1}(U)$ is open in $X$. Moreover, we know by surjectivity of $q$ that $U = qq^{-1}(U)$ and by defition of $\bar{f}$ we get $\bar{f}q =f$. Therefore our question reduces to analyzing when is

$$ \bar{f}(U) = \bar{f}(q(q^{-1}(U)) = f(q^{-1}(U)) $$ open. Once again by surjectivity, we have

$$ f^{-1}(fq^{-1}(U)) = q^{-1}(U) $$

which we assume open in $X$. Therefore, if $f$ were final, this proves that $\bar{f}(U)$ is open for every $U$ open in $X/R$. Reciprocally, if $\bar{f}$ were a homeomorphism, then

$$ V \text{ open in $Y$} \iff \bar{f}^{-1}(V) \text{ open in $X/R$} \iff q^{-1}\bar{f}^{-1}(V) = f^{-1}(V) \text{ open in $X$}. $$

We have finally obtained the following characterization,

Proposition. A map $f : X \to Y$ is a quotient map precisely when the induced map $\bar{f} : X/R \to Y$ is a homeomorphism, where $R$ is the equivalence relation defined by $x \sim y \iff f(x) = f(y)$.


If you're reading Munkres you're making the switch from valuing easiness in your mathematical concepts to valuing power. Your can devise your own analogies using computer programs, woodworking tools, etc.

The Munkres definition will prove to be much more powerful -- stick with it and you'll start to appreciate it, even though sometimes you'll still drop back to the point-wise version, or even good old $\epsilon$-$\delta$.


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