Exercise 18.4

Proof. First, it is easy to see and trivial to formally show that $f$ is injective. The function $f$ can be of course defined as $f(x)=f_1(x)\times f_2(x)$ where $f_1:X\to X$ is the identity function and $f_2:X\to Y$ is the constant function that maps every element of $X$ to $y_0$. Since these have both been proven to be continuous in the text, it follows that $f$ is continuous by Theorem 18.4.

Now let $f^{\prime }$ be the function obtained by restricting the range of $f$ to $f(X)=\left\{x\times y_0\mid x\in X\right\}$. Since $f$ is injective, it follows that $f^{\prime }$ is a bijection. It follows from Theorem 18.2 part (e) that $f^{\prime }$ is continuous. Clearly the inverse function ${f^{\prime }}^{-1}$ is equal to the projection function ${\pi }_1$ so that ${f^{\prime }}^{-1}(x,y)=x$. This was shown to be continuous in the proof of Theorem 18.4. This suffices to show that $f^{\prime }$ is a homeomorphism, which shows that $f$ is an imbedding of $X$ in $X\times Y$. □