Let $X,Y$ be schemes over a field $k$ and $f:X\rightarrow Y$ a morphism. Let us suppose that the fiber $f^{-1}(y)$ of $f$ at a point $y\in Y$ has two connected components $Z_{1},Z_{2}$.
I have read a book in which (in my opinion) the following argument is implicitly used:
This situation cannot happen if $Y$ is smooth, since $y\in f(Z_{1})\cap f(Z_{2})$ would imply that $y$ has double multiplicity in some way.
I would like to know if this holds true under certain conditions, or it is just nonsense.
An easy example: let $X=Spec\left(\mathbb{K}[t,u]_{\displaystyle/(t-1)}\times\mathbb{K}[t,u]_{\displaystyle/(t+1)}\right)=Spec \mathbb{K}[t,u]_{\displaystyle/(t-1)}\coprod Spec\mathbb{K}[t,u]_{\displaystyle/(t+1)}$, let $Y=\mathbb{A}^1_{\mathbb{K}}=Spec\mathbb{K}[t]$ and let $\varphi:X\to Y$ the morphism such that the pull-back $\varphi^{*}$ sends $f\in\mathbb{K}[t]$ in $([f(t,0)],[f(t,0)])\in\mathbb{K}[t,u]_{\displaystyle/(t-1)}\times\mathbb{K}[t,u]_{\displaystyle/(t+1)}$. By definition: every point $y$ of $Y$ is smooth, and $\varphi^{-1}(y)$ has two connected components!
Is it all clear?