Show that the functor $\pi_0:$ TOP $\to$ SET that takes the path components of topological spaces does not preserve pushouts.

Question

Let's say we have the following pushout of topological spaces:

$$
\require{AMScd}
\begin{CD}
A @>{f}>> Y\\
@V{g}VV @VV{G}V\\
X @>{F}>> Z
\end{CD}
$$

and want to show that the following is not necessarily a pushout:

$$
\require{AMScd}
\begin{CD}
\pi_0(A) @>{f_*}>> \pi_0(Y)\\
@V{g_*}VV @VV{G_*}V\\
\pi_0(X) @>{F_*}>> \pi_0(Z)
\end{CD}
$$

One of the arguments I can think of is that there is no obvious reasons that $f_*,g_*,F_*,G_*$ must be such that $F_*g_*(C) = G_*f_*(C)$ for some $C \in \pi_0(A)$. But is this too simplistic?

Another argument could be that even if the second diagram commutes, and if $W$ is a set, and $F_{**}: \pi_0(X) \to W$ and $G_{**}: \pi_0(Y) \to W$ are maps such that

$$
\require{AMScd}
\begin{CD}
\pi_0(A) @>{f_*}>> \pi_0(Y)\\
@V{g_*}VV @VV{G_{**}}V\\
\pi_0(X) @>{F_{**}}>> W
\end{CD}
$$

commutes, there is no unique maps $h: \pi_0(Z) \to W $ such that $F_{**} =h\circ F_{*}$ and $G_{**} =h\circ G_{*}$. But this claim doesn't seem right to me.

Any hints would be greatly appreciated!

Answer 1

Hint: Let $Y$ be the closed topologist's sine curve $\{(x,\sin(1/x)):x\in(0,1]\}\cup\{0\}\times[-1,1]$, with $A=\{0\}\times[-1,1]\subset Y$ the interval on the axis that the sine curve approaches. Can you show that the quotient space $Y/A$ is path-connected?

More details are hidden below.

Let $f:A\to Y$ be the inclusion map and let $X$ be a point, so the pushout is $Y/A$. Since $A$ is path-connected, the pushout after applying $\pi_0$ will just be the same as $\pi_0(Y)$ which has two points. So if we show $Y/A$ is path-connected, then we can conclude that $\pi_0$ fails to preserve this pushout.

To prove that, just note that the first projection $p:Y\to [0,1]$ induces a continuous bijection $Y/A\to[0,1]$, which is a homeomorphism since $Y/A$ is compact and $[0,1]$ is Hausdorff. So $Y/A\cong [0,1]$ and in particular is path-connected.

There are other sorts of examples too; the moral is that quotient topologies are complicated and paths can spontaneously "appear" in a quotient space without coming from the original space. For instance, if you take the quotient of $\mathbb{Q}$ by an equivalence relation with two equivalence classes that are both dense, you get a two-point indiscrete space that is path-connected, even though $\mathbb{Q}$ is totally disconnected. So, writing this quotient as a pushout (which requires a bit of cleverness), $\pi_0$ will not preserve the pushout.