Asking for some advice/verification concerning exercise 5 from chapter 2 of Topology: A Categorical Approach: "Let $\{ X_{\alpha} \}_{\alpha}$ be a collection of spaces, prove that $\pi_0\prod X_{\alpha} \simeq \prod \pi_oX_{\alpha}$." ($\pi_0$ is the functor from $\textbf{Top}$ to $\textbf{Set}$ that takes a space to its set of path connected components.)
My proof: Let $Z$ be a set equipped with the discrete topology. By the universal property of products of topological spaces, for all continuous functions $f_\alpha : Z \rightarrow X_\alpha$, there is a unique map $\varphi: Z \rightarrow \prod X_{\alpha}$ such that $f_\alpha = \pi_\alpha \circ \varphi$:
Now apply $\pi_0$ to the whole diagram. Since $Z$ is totally path disconnected, then $\pi_0 Z = Z$ and any set-function $Z \rightarrow \pi_0 X_\alpha$ can be the image of some map in $\text{Top}(Z,X_\alpha)$ after the functor is applied. In particular, $\pi_o \varphi$ is unique and functors preserve compositions, so we have that for any set-functions $g_\alpha : Z \rightarrow \pi_0X_\alpha$, there is a unique function $\pi_o \varphi : Z \rightarrow \pi_0\prod X_{\alpha}$ such that $g_\alpha = \pi_0 \pi_\alpha \circ \pi_0\varphi$:
In other words, $\pi_0\prod X_{\alpha}$ (together with the corresponding projections $\pi_0 \pi_\alpha$) verifies the universal property of products in $\textbf{Set}$. Thus, $\pi_0\prod X_{\alpha} \simeq \prod \pi_oX_{\alpha}$ as desired.
I'm not quite confident with this proof since I didn't end up using any of the maps that define paths (the ones from $[0,1]$), which seems a bit too good to be true. I'm also not sure that the "trick" of using a space with the discrete topology works the way I described; I'm not quite comfortable with functors just yet. Any advice/corrections/alternative proofs are welcome, thanks!