# Path connectedness is a homotopy invariant

I haven’t updated this blog in a while, due to qualifying exam preparation and research projects. I’ll write about the latter soon.  While studying for the geometry/topology qual, I asked a basic question: Is path connectedness a homotopy invariant? Turns out the answer is yes, and I’ve written up a quick proof of the fact below.  You can view a pdf of this entry here.

Proposition 1
Let ${f : X \rightarrow Y}$ be a homotopy equivalence. If ${X}$ is path connected, then so is ${Y}$.

Proof: It’s clear that the image of ${f}$ is path connected. Thus, it is enough to show that any point of ${Y}$ can be connected to a point of ${f(X)}$. Let ${g : Y \rightarrow X}$ be a map such that ${f\circ g}$ is homotopic to ${\text{id}_Y}$, via the homotopy ${h : Y \times I \rightarrow Y}$. Let ${y \in Y}$. Then, ${y' = f(g(y)) \in f(X)}$ and ${\gamma(t) = h(y, t)}$ is a path from ${y'}$ to ${y}$. $\Box$

We can deduce the following result from the proof of proposition 1.

Corollary 2 Let ${f : X \rightarrow Y}$ be a homotopy equivalence. Then ${X}$ is path connected if, and only if, ${f(X)}$ is path connected.

Proof: If ${f(X)}$ is path connected, then ${Y}$ is path connected. Hence ${X}$ is path connected. $\Box$