The coordinate axes form a reduced crossing

The germs $x$ and $y$ are irreducible. Neither lies in ${\mathfrak{m}}_0^2$, because both have nonzero linear part; if $x=ab$ with $a,b$ nonunits, then $a,b\in {\mathfrak{m}}_0$ by [F3] and hence $x=ab\in {\mathfrak{m}}_0^2$, a contradiction. So $x$ is not a product of two nonunits, and the same argument applies to $y$. The two germs are not associates: $x=uy$ with $u$ a unit would give $x(t,0)=u(t,0)\cdot 0=0$ for every small $t$, contradicting $x(t,0)=t\mathrm{\ne }0$ for $t\mathrm{\ne }0$. Hence $f=xy$ is a product of two pairwise nonassociate irreducibles, each occurring once, so $f$ is reduced and its reduced defining germ is $f$ itself, with $I_0(X)=(f)$ by [F1].

Let $\mathrm{\Delta }\subseteq \mathbb{C}$ be a connected open set containing the origin and let $\gamma =(u,v):\mathrm{\Delta }\to {\mathbb{C}}^2$ be holomorphic with $\gamma (\mathrm{\Delta })\subseteq X$, that is, $u(t)v(t)=0$ for every $t\in \mathrm{\Delta }$. By [F6] applied to the connected domain $\mathrm{\Delta }$, one of the two coordinate functions vanishes identically, so the image of $\gamma$ is contained in a single axis: either $\gamma (\mathrm{\Delta })\subseteq Z(x)$ or $\gamma (\mathrm{\Delta })\subseteq Z(y)$.

By [F4] applied to the factorisation $f=x\cdot y$ of step 1.1, $X=Z(x)\cup Z(y)$, and the two branches $Z(x)=\{x=0\}$ and $Z(y)=\{y=0\}$ are exactly the irreducible components of $X$; they are distinct as set germs and neither contains the other.

Every point of $Z(x)$ is regular: $Z(x)$ is the graph $\{(x,y):x=0\}$ of the zero function over the $y$-coordinate near each of its points, hence a holomorphic hypersurface graph near every such point, so [F5] gives regularity. The same argument exhibits $Z(y)=\{(x,y):y=0\}$ as the graph of the zero function over the $x$-coordinate, so every point of $Z(y)$ is regular as well.

The origin is a singular point: by step 1.1 the reduced defining germ of $X$ is $xy$, and $d(xy)=ydx+xdy$ vanishes at $0$. So $0$ is not regular by [F5]. Since $Z(x)\cap Z(y)=\{0\}$, every point of $X$ other than the origin lies on exactly one of the two branches and is regular by step 3.1; hence the origin is the only singular point of $X$ in a neighbourhood of $0$, and it is exactly the crossing of the two branches.

Suppose first that $\gamma (\mathrm{\Delta })\subseteq Z(x)$. The set germ of the image of $\gamma$ at the origin is then contained in $Z(x)$, which is a proper subgerm of $X$: for every small $\epsilon \mathrm{\ne }0$ the point $(\epsilon ,0)$ belongs to $X$ but not to $Z(x)$. Hence the image of $\gamma$ cannot contain a full representative of $X$, so its image germ is not $X$; the case $\gamma (\mathrm{\Delta })\subseteq Z(y)$ is the same with the roles of $x$ and $y$ exchanged. Therefore no holomorphic map of a connected disc has image germ $X$, injective or not, and in particular no single injective branch parametrisation covers both components. The individual branches are parametrised by the injective holomorphic maps $t\mapsto (t,0)$ and $t\mapsto (0,t)$, whose images are full representatives of $Z(y)$ and $Z(x)$ respectively.