Email by Lenny Taelman dated May 1, 2016.

Remark 75.8.3. A universal homeomorphism of algebraic spaces need not be representable, see Morphisms of Spaces, Example 66.53.3. In fact Theorem 75.8.1 does not hold for universal homeomorphisms. To see this, let $k$ be an algebraically closed field of characteristic $0$ and let

$\mathbf{A}^1 \to X \to \mathbf{A}^1$

be as in Morphisms of Spaces, Example 66.53.3. Recall that the first morphism is étale and identifies $t$ with $-t$ for $t \in \mathbf{A}^1_ k \setminus \{ 0\}$ and that the second morphism is our universal homeomorphism. Since $\mathbf{A}^1_ k$ has no nontrivial connected finite étale coverings (because $k$ is algebraically closed of characteristic zero; details omitted), it suffices to construct a nontrivial connected finite étale covering $Y \to X$. To do this, let $Y$ be the affine line with zero doubled (Schemes, Example 26.14.3). Then $Y = Y_1 \cup Y_2$ with $Y_ i = \mathbf{A}^1_ k$ glued along $\mathbf{A}^1_ k \setminus \{ 0\}$. To define the morphism $Y \to X$ we use the morphisms

$Y_1 \xrightarrow {1} \mathbf{A}^1_ k \to X \quad \text{and}\quad Y_2 \xrightarrow {-1} \mathbf{A}^1_ k \to X.$

These glue over $Y_1 \cap Y_2$ by the construction of $X$ and hence define a morphism $Y \to X$. In fact, we claim that

$\xymatrix{ Y \ar[d] & Y_1 \amalg Y_2 \ar[l] \ar[d] \\ X & \mathbf{A}^1_ k \ar[l] }$

is a cartesian square. We omit the details; you can use for example Groupoids, Lemma 39.20.7. Since $\mathbf{A}^1_ k \to X$ is étale and surjective, this proves that $Y \to X$ is finite étale of degree $2$ which gives the desired example.

More simply, you can argue as follows. The scheme $Y$ has a free action of the group $G = \{ +1, -1\}$ where $-1$ acts by swapping $Y_1$ and $Y_2$ and changing the sign of the coordinate. Then $X = Y/G$ (see Spaces, Definition 64.14.4) and hence $Y \to X$ is finite étale. You can also show directly that there exists a universal homeomorphism $X \to \mathbf{A}^1_ k$ by using $t \mapsto t^2$ on affine spaces. In fact, this $X$ is the same as the $X$ above.

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