Reduction of graph chromatic number to hypergraph 2-colorability I'm following this paper titled "Coverings and colorings of hypergraphs" by Lovasz 1973, which is referenced in Garey and Johnson's Computers and Intractability, for the Set Splitting Problem. In this paper, the author tries to reduce graph chromatic number to hypergraph 2-colorability (same as set splitting). 

You can find the paper here: http://web.cs.elte.hu/~lovasz/old-papers.html

I'm reproducing the reduction below.

Let $G$ be a graph, $V(G) = \{x_1,\dots,x_n\}$. Let $G_i$ be an isomorphic copy of $G$, $(i=1,\dots,k)$, $V(G_i) = \{x_{i,1},\dots,x_{i,n}\}$ ($x_{iv}$ is a point corresponding to $x_v$). Take a new point $y$ and let $f_v = \{x_{1,v},\dots,x_{k,v},y\}$. Define hypergraph $H$ by
$$
  H = E(G_1) \cup \dots \cup E(G_k) \cup \{f_1,\dots,f_n\}.
$$
Then $H$ is 2-colorable if and only if $G$ is $k$-colorable. I don't understand this conclusion and the only example I've been able to come up with where this is true is $k = 1$.
As the other answer points out, the reduction in the original paper seems to have a bug: $H$ will not be two-colorable unless $G$ is bipartite. I couldn't quite see how to prove the reduction in the other answer works, but here is one that I think does.

The vertices of $H$ will be $\{x_{i, v}: i \in [k], v \in V(G)\} \cup \{z\}$. $H$ has an edge $f_v = \{x_{1,v}, \ldots, x_{k,v}\}$ for every $v \in V(G)$,  and, for each $(u,v) \in E(G)$, and each $i \in [k]$, an edge $\{z, x_{i,u}, x_{i,v}\}$.

Given a coloring $\chi_G:V(G) \to [k]$ of $G$ you can construct a $2$-coloring $\chi_H:V(H) \to \{0,1\}$ of $H$ by setting 

$$
\chi_H(z) = 1,\\
\chi_H(x_{i,v}) = 1 \iff \chi_G(v) = i.
$$

It's easy to check that no edge of $H$ is monochromatic.

In the other direction let $\chi_H$ be a proper 2-coloring of $H$, and assume that $\chi_H(z) = 1$. (If this is not the case just rename the colors.) Since no edge $f_v$ is monochromatic, for every $v \in V(G)$ there is some $i$ such that $\chi_H(x_{i,v}) = 1$. Pick an arbitrary such $i$ and set $\chi_G(v) = i$. This is a proper coloring, or otherwise there would be an $i \in [k]$ and an edge $(u,v) \in E(G)$ such that $\chi_H(x_{i,u}) = \chi_H(x_{i,v}) = 1$, which would make the edge $\{z, x_{i,u}, x_{i,v}\}$ of $H$ monochromatic.