$f$ is a local homeomorphism iff it is open with open diagonal Let $f$ be a continuous map $f : X \rightarrow Y$ . Then $f$ is a local homeomorphism if and only if $f$ is open and the diagonal of $f$ which is $d: X\rightarrow X×_YX$ is open.
Iam really stuck with this problem..... please help.....
 A: Just for clarity, we define $X \times_Y X = \{ (x , x') \in X \times X : f(x) = f(x') \}$ and give it the subspace topology as a subset of $X \times X$. Furthermore, define $d : X \to X \times_Y X$ to be the map sending $x \mapsto (x , x)$.
($\Rightarrow$). Suppose that $f : X \to Y$ is a local homeomorphism. Given any open $U \subseteq X$ and any $x \in U$, because $f$ is a local homeomorphism, there exists an open set $V$ containing $x$ such that $f(V) \subseteq Y$ is open and $f|_V : V \to f(V)$ is a homeomorphism. Then $U \cap V$ is open in $V$ so $f(U \cap V)$ is open in $f(V)$, and therefore in $Y$. Furthermore, $f|_{U \cap V} : U \cap V \to f(U \cap V)$ is also a homeomorphism. Hence the set
\begin{equation*}
\mathcal{B} = \{ W \subseteq X : W \subseteq X \text{ and } f(W) \subseteq Y \text{ are open and } f|_W : W \to f(W) \text{ is a homeomorphism} \}
\end{equation*}
is a basis for the topology on $X$.

*

*($f$ is open). Let $U \subseteq X$ be open, and let $y \in f(U)$. Then there exists $x \in U$ such that $f(x) = y$. Let $W \in \mathcal{B}$ be such that $x \in W \subseteq U$. Then $y \in f(W) \subseteq f(U)$, and $f(W)$ is open in $Y$. Since $y \in f(U)$ was arbitrary, it follows that $f(U)$ is open in $Y$.

*($\Delta_f$ is open). Let $U \subseteq X$ be open, and let $(x , x) \in d(U)$, so that $x \in U$. As above, there exists a set $W \in \mathcal{B}$ such that $x \in W \subseteq U$. Then $W \times W$ is open in $X \times X$, and therefore $(W \times W) \cap (X \times_Y X)$ is open in $X \times_Y X$. However we claim that:
\begin{equation*}
(W \times W) \cap (X \times_Y X) \subseteq d(U)
\end{equation*}
To see this, take any $(z , z') \in (W \times W) \cap (X \times_Y X)$. On the one hand, $(z , z') \in W \times W$, so $z , z' \in W \subseteq U$, and on the other hand $(z , z') \in X \times_Y X$, hence $f(z) = f(z')$, but because $f|_W$ is a homeomorphism onto its image, it is injective, and hence $z = z'$, so $(z , z') = (z , z) = d(x) \in d(U)$.
Hence $(x , x) \in (W \times W) \cap (X \times_Y X) \subseteq d(U)$ and $(W \times W) \cap (X \times_Y X)$ is open in $X \times_Y X$. Since $(x , x) \in d(U)$ was arbitrary, it follows that $d(U)$ is open in $X \times_Y X$.

($\Leftarrow$). Suppose $f$ and $d$ are open maps. Let $x \in X$. Since $d$ is an open map, $d(X)$ is open in $X \times_f X$ in the subspace topology. Therefore there exist open sets $U , V \subseteq X$ such that $(x , x) \in (U \times V) \cap (X \times_Y X) \subseteq d(X)$. We claim that $f|_{U \cap V}$ is injective. To see this, suppose $z , z' \in U \cap V$ are such that $f(z) = f(z')$, then $(z , z') \in (U \times V) \cap (X \times_Y X)$, and hence $(z , z') \in d(X)$, so $z = z'$, as desired. Furthermore, $U \cap V$ is open in $X$, so since $f$ is an open map, $f(U \cap V)$ is open in $Y$. Finally, the restriction of an open map to an open set is still an open map, hence $f|_{U \cap V}: U \cap V \to f(U \cap V)$ is continuous, open, and bijective, and hence is a homeomorphism.
