Given a product topology X = $\prod_{i \in I}X_i$ , check: if E is closed in X, then $\pi_i(E)$ is closed in $X_i$

Question

Given: A family of topological spaces: $\{(X_i, \mathcal{T_i})\mid i \in I\}$

$X = \prod_{i \in I}X_i$ is a product topology with topology defined as $\mathcal{T}$

$\pi_i: X \rightarrow X_i$ is projection mapping

Conjecture: If $E$ is closed in $X$, then $\pi_i(E)$ is closed in $X_i$. True or False?

Here's what I was thinking:

Given projecting mapping, $\pi_i(\cdot)$ is continuous by construction.

By hypothesis, $E$ is closed in $X,$ then $X-E$ is open in $X.$

Let $U=X-E$. Take any $x \in U$, $\pi_i(x) \in X_i$

Take an open set $V \in X_i$ such that $\pi_i(x) \in V$

Since $\pi_i(\cdot)$ is continuous, $\pi(U(x)) \subset V(\pi_i(x))$

Take $W = X_i - V(\pi_i(x))$ $\implies$ $W$ is closed

... Something along these lines

While I have seen counter examples, I wish to know what is wrong with my solution.

Answer 1

The graph of $y=e^x$ is closed, but its projection onto the $y$ axis is not closed.