It is said in wikipedia, nlab and Terilla that subspace has a universal property.

Universal property for the subspace topology. For every topological space $(Z, \tau_Z)$ and every function $f : Z \to Y$, $f$ is continuous if and only if $i \circ f : Z \to X$ is continuous.

Here's a picture

$$ \begin{array}{ccc} {} & {} & X \\ {} & \overset{i \circ f}{\nearrow} & \uparrow i \\ Z & \underset{f}{\to} & Y \end{array} $$ One should think of the universal property stated above as a property that may be attributed to a topology on $Y$. At this point, you may think that some topologies have this property and some do not. Theorem $1$ means that the subspace topology on $Y$, as previously defined, does have this universal property. Furthermore, the subspace topology is the only topology on $Y$ with this property. Let's prove it.

But, as I know, universal property is some sort of 'initial' or 'final' property in another category such as object: $X\xrightarrow f Y, \forall Y, f$ and morphism $\sigma: Y\rightarrow Z$ makes the triangle $X\xrightarrow f Y$ and $X\xrightarrow g Z$ commute, thats to say $g = \sigma f$. However, I cannot see any universal property in this triangle, since $i\circ f$ is not 'any' morphism, but have connections with what isn't known yet - the morphism $f$.

So, how can we understand this as a universal property ? Or, why a universal property can have connection with 'f'?

  • 1
    $\begingroup$ Please do not use images to convey information not otherwise present in your post. $\endgroup$ Commented Jul 27, 2023 at 3:15
  • 4
    $\begingroup$ Welcome to mse! I've edited your question to use mathjax (which is searchable) rather than an image (which isn't) so that other users of the site will have an easier time finding this question. In the future you should do the same ^_^ $\endgroup$ Commented Jul 27, 2023 at 3:17
  • 1
    $\begingroup$ @HallaSurvivor Thanks a lot! I will do the same in my following post! Sorry for the late thank you, for I'm not sure whether it's OK say thank you in comment :( $\endgroup$
    – Liam
    Commented Apr 6 at 6:51
  • 1
    $\begingroup$ @Liam -- No worries at all! Comments are usually quite conversational. You would need to encounter someone particularly rude for them to tell you not to thank someone in the comments! I've left comments with even less content than that, haha $\endgroup$ Commented Apr 6 at 17:06

2 Answers 2


This universal property is to mediate between arbitrary functions (defined on the underlying set of some space) and continuous functions (which make reference to the topology). This should make some sense, since we're already given $Y$ (and thus, access to functions into $Y$) and we're trying to choose the right topology to equip $Y$ with.

Here's another place you may have seen this idea: We know the product of $(X,\tau_X)$ and $(Y, \tau_Y)$ must have $X \times Y$ as its underlying set. But how can we tell whether a function $f : (Z, \tau_Z) \to X \times Y$ "should" be continuous? Well we say that $f$ is continuous if and only if both $\pi_X \circ f : (Z, \tau_Z) \to (X, \tau_X)$ and $\pi_Y \circ f : (Z, \tau_Z) \to (Y, \tau_Y)$ are continuous! So you see, we've been given a function $f$ on underlying sets, and the universal property tells us whether or not $f$ is continuous.

In your example, we have a similar question: How can we tell when an arbitrary function $f : (Z, \tau_Z) \to Y \ (\subseteq X)$ should be continuous? Well, we want $f$ to be continuous if and only if $\iota \circ f : (Z, \tau_Z) \to (X, \tau_X)$ is!

I agree that these don't look like the kinds of simple universal properties that you might be used to. That's basically because (as I said at the start of this answer) they're universal properties mediating which maps in $\mathsf{Set}$ lift to maps in $\mathsf{Top}$, rather than the usual universal properties which govern which maps we can construct at all. But we can still formulate these as universal properties that you might recognize (it just takes some work). If you like, there is a lattice of possible topologies you can put on $Y$, and the universal property is saying that there's an initial/terminal object in this lattice of possible topologies (viewed as a thin category).

This can be made precise with the language of "lifts of $U$-structured sources/sinks", central to the definition of a topologically concrete category (that is, a category whose relationship to $\mathsf{Set}$ is structurally similar to the relationship between $\mathsf{Top}$ and $\mathsf{Set}$). You can read more about these ideas in Adamek, Herrlich, and Strecker's The Joy of Cats.

I hope this helps ^_^


As its name indicates, the given definition indeed does not define an object with a universal property in the category of topological spaces. Any such object would only be well-defined up to homeomorphism, whereras “Y with some topology on it” clearly does not have that kind of flexibilty. Here’s my attempt to vary the definition in a way that yields an object with a universal property in $\mathrm{Top}$, the category of topological spaces.

Let’s say the space defined by $i\colon Y \to X$ is a topological space $\underline{Y}$ together with a map of sets $y\colon \underline{Y}\to Y$ such that $i\circ y$ is continuous, and such that the following universal property holds: For every map of sets $f\colon Z\to Y$ such that $i\circ f$ is continuous, there exists a unique continuous map $\underline{f}\colon Z \to \underline{Y}$ such that $y\circ \underline{f} = f$.

\begin{array}{ccc} {} & {} & X\;\; \\ {} & \overset{\text{cont.}}{\nearrow} & \uparrow i \\ Z & \underset{\forall f}{\to} & Y\;\; \\ {} & \underset{\exists !\underline{f}}{\searrow} & \uparrow{y} \\ & & \underline{Y}\;\; \end{array}

This looks more complicated, but on the plus side it looks very similar to the categorical definition of a kernel in, say, abelian groups:

\begin{array}{ccc} {} & {} & B\;\; \\ {} & \overset{0}{\nearrow} & \uparrow g \\ Z & \underset{\forall f}{\to} & A\;\; \\ {} & \underset{\exists !\underline{f}}{\searrow} & \uparrow \\ & & \mathrm{ker}(g)\;\; \end{array}

This is a universal property: Formally, such a pair $(\underline{Y},y)$ represents the functor $$ F\colon \mathrm{Top}^{\mathrm{op}} \to \mathrm{Sets} $$ given by $F(Z) := \{ g: Z → Y \text{ in } \mathrm{Sets} \text{ such that $i\circ g$ is continuous}\}$ on objects and in the obvious way on morphisms. That is, $y\in F(\underline{Y})$, and the natural transformation of functors $\mathrm{Hom}_{\mathrm{Top}}(-,\underline{Y}) \to F$ defined by $y$ (namely $\underline{f}\mapsto (F\underline{f})(y) = y \circ \underline{f}$) is a natural isomorphism.

Relation to usual definition: Taking $Z$ to be a one-point space, we see immediately that $y$ needs to be a bijection. So we might as well take $y=\mathrm{id}$, and then the diagram simplifies to the usual diagram defining the subspace topology, as in the question.

Relation to sinks and sources: [EDIT] I realize in hindsight that I have simply spelled out the definition of an initial lift of the sturctured source $Y\to U(X)$, as alluded to in Chris Grossack's answer. Here, $U\colon \mathrm{Top} \to \mathrm{Sets}$ denotes the forgetful functor.

  • $\begingroup$ nice answer ... +1 $\endgroup$
    – TShiong
    Commented Apr 30 at 21:33
  • $\begingroup$ Thanks! Nice answer. $\endgroup$
    – Liam
    Commented May 2 at 1:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .