# Is “connected, simply connected” Redundant?

Here are my definitions of "connected" and "simply connected."

A topological space $X$ is connected if and only if it is not the union of two nonempty disjoint open sets.

A topological space $X$ is simply connected if and only if it is path-connected and has trivial fundamental group (i.e. $\pi_1(X)\simeq\{\mathrm{e}\}$ and $|\pi_0(X)|=1$).

It is a classic and elementary exercise in topology to show that, if a space is path-connected, then it is connected. Thus, if a space is simply connected, then it is connected.

Yet, despite this implication, I've read several cases where the words "connected, simply connected" appear together.

For example, in Kobayashi and Nomizu's Foundations of Differential Geometry, Volume 1, page 252, the following is written: "Let $M$ be a connected, simply connected analytic manifold with an analytic linear connection."

In this paper on $T^3$ actions and this paper on rotationally symmetric manifolds (Links to Journal Storage, also known as JSTOR), they use "connected, simply connected" on their first page. The first of these takes it a step further, referring to "closed, compact, connected, simply connected $4$-manifolds," where a closed manifold (unfortunate terminology) is defined as a compact manifold without boundary.

My question is simple: Why use the words "connected, simply connected" when being simply connected implies connectedness?

Thank you.

• I've seen people use "simply connected" to mean that every connected component has trivial fundamental group [don't remember who and where, unfortunately]. In that usage, it isn't redundant. – Daniel Fischer Sep 18 '14 at 16:20
• Interesting. I would think a term like "component-wise simply connected" would work better. That usage would make sense, though. – Robin Goodfellow Sep 18 '14 at 16:22
• This is an example of unnecessary redundant overdetermination. I do it myself. – Lee Mosher Sep 18 '14 at 23:00
• The idea of the fundamental group or the entire homotopy theory is intrinsic to point topological category. Thus $$\pi_{1} (X)$$ is dependant on base point. – Illuminata Nov 14 '17 at 4:26