# What is the use of locally connected spaces?

One of the main properties of locally connected spaces is that their connected components are clopen and thus, they are homeomorphic to the colimit of their connected components.

This is good to study the $\Pi_0$ functor, it gets a right adjoint.

At the topos level, a topos is declared locally connected precisely if we have such an adjunction (or equivalently if every object is isomorphic to the coproduct of connected objects).

But, being locally connected is more than just having open connected components (see all the classical connected but not locally connected spaces).

My questions are : why are people interested in locally connected spaces and not in spaces with open connected components ?

Is there a full sub-category of $\textrm{Top}$ with all connected and locally connected spaces on which $\Pi_0$ has a right adjoint ?

• Your queestion is backwards, really. Lots of spaces in which people are interested happen to be locally connected and that property is useful on occasion, so we abstract it and study locally connected spaces... – Mariano Suárez-Álvarez Sep 28 '13 at 20:16
• In fact, I would say that it takes some inventiveness to come up with a space which is not locally connected... – Mariano Suárez-Álvarez Sep 28 '13 at 20:17
• I'd like to see a good answer to this question, but it seems to me that at least in elementary algebraic topology there is not that much emphasis put to locally connected spaces. This is too weak a condition to develop covering theory and later on it is often very conveniant to assume that your space is a CW-complex. – Piotr Pstrągowski Sep 30 '13 at 17:04
• Good question, deserves more love. – goblin Oct 4 '15 at 10:42
• @goblin could you take a look at my answer and make sure I haven't written nonsense? – Arrow May 29 '16 at 22:23

From a categorical perspective, the subcategory of $\mathsf{Top}$ generated by spaces with open connected components does not have pullbacks.

Borceux and Janelidze, Galois Theories, section 6.4:

The category $\mathsf{Top}$ of topological spaces is extensive, but not of the form $\mathsf{Fam}(\mathcal A)$. On the other hand the category of topological spaces with open connected components (see proposition 6.1.1), which is the "obvious best replacement" of $\mathsf{Top}$ by a category of the form $\mathsf{Fam}(\mathcal A)$, does not have pullbacks.

Regarding your second question (there's probably a mistake somewhere, since I doubt nobody else would have thought of this before):

First of all, I think $\Pi_0$ has a right adjoint whenever it exists and $\mathcal A$ admits a terminal object. It is given by the discrete functor $H$ defined by $A\mapsto A\cdot \mathbf{1}=\coprod_A\mathbf{1}$. The very existence of a connected components functor means your category is actually of the form $\mathsf{Fam}(\mathcal A)$ i.e is a free coproduct cocompletion. These are precisely the extensive categories in which every object has a presentation as a coproduct of connected objects - the connected components.

Now, I think that for a space $X$ TFAE:

• $X$ is the coproduct of its connected components.
• $X$ has open connected components.

Proof. The fact $\mathsf{Top}$ is extensive ensures a coproduct decomposition is unique up to homeomorphism. By definition a space is disconnected if it has a nontrivial coproduct presentation by open subspaces. Hence, the uniqueness up to isomorphism guarantees that if $X$ is the coproduct of its connected components, they are open. You already wrote the converse.

This means the largest full subcategory of $\mathsf{Top}$ on which $\Pi_0$ is defined is the category of spaces with open connected components.

Added. In any category there is a notion of connected object. This notion is especially well behaved in extensive categories, in which coproducts interact well with pullback. In a "spatial" category, ideally, every object should have a presentation as a unique up to iso coproduct of connected objects - a canonical decomposition into simple parts. However, this is often not true - totally disconnected spaces need not be discrete and so are not the coproducts of copies of $\mathbf{1}$.

Given a category $\mathcal A$, $\mathsf{Fam}(\mathcal A)$ is its free coproduct completion. What it does is freely append coproducts of connected objects. It can be defined as a fibration, as in section 6.1 of Galois Theories

For a family $(A_i)_{i\in I}$ of objects of a category $\mathcal A$ we write $$(A_i)_{i\in I}=A=(A_i)_{i\in I(A)}$$ considering $I$ as a functor $$I:\mathsf{Fam}(\mathcal A)\longrightarrow \mathsf{Set}$$ from the category of all families of objects in $\mathcal A$ to the category of sets.

In this way, (writing $\Pi_0=I$) every object $(A_i)_{i\in \Pi_0(A)}$ in $\mathsf{Fam}(\mathcal A)$ is identified as the collection of connected components of the object $\coprod_{i\in \Pi_0(A)}A_i$.

The existence of $\Pi_0$ is the equivalent to saying that every object has a unique up to iso presentation as a coproduct of connected objects, and that's why it's important.

• It would be nice if you could give a vague, imprecise, birds-eye overview of what $\mathsf{Fam}(-)$ is and why its interesting. – goblin May 30 '16 at 6:45
• @goblin tried to do that :) – Arrow May 30 '16 at 8:21