Statement
Let be $A$, $B$, $C$ and $D$ topological spaces and let be $\phi:A\rightarrow C$ and $\psi:B\rightarrow D$ two continuous function. So the product function $\Delta:A\times B\rightarrow C\times D$ defined through the condition
$$
\Delta(a,b):=\big(\phi(a),\psi(b)\big)
$$
for any $(a,b)\in A\times B$ is continuous in the product topology.
Clearly $\pi_A\big(\Delta(a,b)\big)=\phi(a)$ and $\pi_B\big(\Delta(a,b)\big)=\psi(b)$ but $\pi_A\circ\Delta: A\times B\rightarrow A$ and $\pi_B\circ\Delta:A\times B\rightarrow D$ whereas $\phi:A\rightarrow C$ and $\psi: B\rightarrow D$ so I think that I can't use the universal mapping theorem for products to claim that $\Delta$ is continuous. So could someone help me, please?
Best Answer
Here's a proof by abstract nonsense:
Consider the compositions $\phi\circ\pi_A:A\times B\to C$ and $\psi\circ \pi_B:A\times B\to D$. These are continuous, so by the universal property of $C\times D$ we have a unique continuous map $\Gamma:A\times B\to C\times D$ making the obvious diagram commute, i.e. satisfying $\pi_C\circ \Gamma=\phi\circ \pi_A$ and $\pi_D\circ \Gamma=\psi\circ \pi_B$. Since we also have $\pi_C\circ \Delta=\phi\circ \pi_A$ and $\pi_D\circ \Delta=\psi\circ \pi_B$, it follows that $\Delta=\Gamma$. Therefore, $\Delta$ is continuous.