Given \(x_0 \in X\) and \(y_0 \in Y\), show that the maps \(f: X \to X \times Y\) and \(g: Y \to X \times Y\) defined by
\[ f(x) = x \times y_0 \quad \text{and} \quad g(y) = x_0 \times y \]
are imbeddings.
Solution
An imbedding is a map that is a homeomorphism onto its image.
1. \(f\) is continuous and injective.
Its coordinates are \(\pi_1\circ f=\operatorname{id}_X\) and the constant map \(\pi_2\circ f\equiv y_0\), both continuous, and a map into a product is continuous if and only if its coordinates are. If \(f(x_1)=f(x_2)\), then \((x_1,y_0)=(x_2,y_0)\), so \(x_1=x_2\).
2. \(f\) is a homeomorphism onto \(X\times\{y_0\}\).
The inverse of \(f\colon X\to X\times\{y_0\}\) is the restriction of \(\pi_1\) to \(X\times\{y_0\}\), which is continuous.
3. \(g\) is an imbedding.
Exchange the roles of the factors in steps 1 and 2: \(g\) has coordinates the constant map at \(x_0\) and \(\operatorname{id}_Y\), and its inverse on \(\{x_0\}\times Y\) is the restriction of \(\pi_2\).
4. Q.E.D.
Steps 1 and 2 show that \(f\) is an imbedding, and step 3 treats \(g\).