Limits in the category of topological spaces must come equipped with the initial topology

Question

$ \newcommand{\cat}[1]{\mathsf{#1}} $The title says it all. Let $ \cat J $ be a small category and let $ D $ be a $ \cat J $-indexed diagram of topological spaces. Suppose a limit $ (L,\tau_L) $ of this diagram exists in the category of topological spaces (it exists, indeed, but that's not important here), and let $ {\left(\lambda_j\colon (L,\tau_L)\to Dj\right)}_{j\in \cat J} $ be (the legs of) its corresponding limit cone.

Is it true that $ \tau_L $ is the initial topology on the set $ L $ with respect to the continuous maps $ \lambda_j\colon (L,\tau_L)\to Dj $?

What I tried so far

I tinkered with products just to get a feel for what I'm doing.

Let $ (A,\tau_A) $ and $ (B,\tau_B) $ be topological spaces. Let $ (P,\tau) $ be a categorical product of $ (A,\tau_A) $ and $ (B,\tau_B) $, and let $ \pi_A\colon (P,\tau)\to (A,\tau_A) $ and $ \pi_B\colon (P,\tau)\to (B,\tau_B) $ be (the legs of) its limit cone. Suppose $ \tau^\prime $ is a topology on the set $ P $ such that the set maps $ \pi_A\colon P\to A $ and $ \pi_B\colon P\to B $ upgrade to continuous functions $ \pi_A\colon (P,\tau^\prime)\to (A,\tau_A) $ and $ \pi_B\colon (P,\tau^\prime)\to (B,\tau_B) $ (I'm using the same names for the projections everywhere just to confuse you, or just because naming them differently is worse than the disease).

I need to prove that given $ U\in \tau $, then $ U\in \tau^\prime $. I thought that I could "pull back" $ U $ from $ (P,\tau) $ to $ (P,\tau^\prime) $ by the means of the continuous function $ \phi\colon (P,\tau^\prime)\to (P,\tau) $ whose existence is guaranteed by the universal property of products. If we are to follow this line of reasoning, we could well hope not only that $ \phi^{-1}(U) = U $, but that $ \phi $ has the identity function of $ P $ as the underlying set mapping.

From there I'm basically lost. Showing that the identity function $ 1_P\colon P\to P $ give rise to a continuous function $ 1_P\colon (P,\tau^\prime)\to (P,\tau) $ is the same as showing that $ \tau $ is coarser than $ \tau^\prime $, which is what I wanted to prove.

Answer 1

You're exactly right, and there's a few ways to see this.

Probably the simplest way is to use the "characteristic property" of the initial topology, which you can find on the relevant wikipedia page. This says that if $(f_i : X \to Y_i)$ is a family of maps, then the initial topology is such that $g : Z \to X$ is continuous if and only if each of the $f_i \circ g$ are. If your definition of initial topology is in terms of open sets rather than characteristic properties, you can find a proof that the two definitions are equivalent here.

Now, we know that the forgetful functor $U : \mathsf{Top} \to \mathsf{Set}$ preserves limits (after all, it is right adjoint to the discrete topology functor) so if we have some diagram $D$, we know that $U (\lim_D Y_i) \cong \lim_D (U Y_i)$, and the underlying set is totally determined. All that's left is to choose the "correct" topology to put on this set, of course the "characteristic property" as defined above tells us exactly that the initial topology is the one to choose!

Can you run a similar argument to show that colimits in $\mathsf{Top}$ are computed as colimits in $\mathsf{Set}$ equipped with the final topology? The idea here is to use that $U$ also has a left adjoint (the indiscrete topology functor), so $U$ preserves colimits too!

As a last aside, there is a notion of a "topologically concrete category" which axiomatizes the relationship $\mathsf{Top}$ has to $\mathsf{Set}$. The really important part of the definition is the existance of "initial lifts", which basically tell you that initial topologies exist, and they're what we use to compute limits! You can see some examples of topological categories, and their similarities to $\mathsf{Top}$ in a blog post of mine.

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I hope this helps ^_^

Answer 2

In this answer I showed that initial topologies obey the so-called universal theorem on continuity (if $\tau$ on $X$ is initial for the functions $f_i: X\to (Y_i,\tau_i), i \in I$ then any function $g:(Z,\tau') \to (X,\tau)$ into it, is continuous iff all compositions $f_i \circ g$ are continuous) characterises the initial topology $\tau$ (the topology $\tau$ on $X$ is the unique topology on $X$ with that property). The argument is not hard.

So if we have a diagram in the category $\mathsf{Top}$, the same diagram in $\mathsf{Set}$ (so applying the forgetful functor) has a limit $X$ (plus maps) in $\mathsf{Set}$ and as that limit must obey the universal theorem for continuity back in $\mathsf{Top}$ (the function as set map exists and we must make it exist, i.e. continuous), $X$ must get the initial topology by the previous fact. A bit handwavy, but I think a seasoned category theorist could make it a formal one (we sort of have a lift from $\mathsf{Set}$ back to $\mathsf{Top}$. I think Herrlich, Preuß and Strecker (among others) have writen a lot about such "Topological categories", which for them also included the categories of uniform spaces, measurable spaces, convergence spaces and closure spaces, as well as proximity spaces.In all of these we have an equivalence of initial structures (often called "sources") and final structures ("sinks") to define limits.