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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?

Antonio Maria Di Mauro
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    Hint: An open set in the product topology looks like $U\times V$, where $U$ and $V$ are open in their respective topological spaces. Show that $\Delta^{-1}(U\times V)$ is open in $A\times B$. – boink Jun 14 '20 at 18:15
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    @boink Every open set in the product topology is a union of sets of the form $U\times V$. – Reveillark Jun 14 '20 at 19:07

3 Answers3

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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.

Reveillark
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  • Would you present a topological proof? – William Elliot Jun 14 '20 at 19:03
  • @WilliamElliot The topological proof is essentially outlined in boink 's comment. It suffices to look at basic open sets in $C\times D$, so fix $U$ and $V$ open in $C$ and $D$, respectively. Then $\Delta^{-1}(U\times V)=\phi^{-1}(U)\times\psi^{-1}(D)$, which is open in $A\times B$ because $\phi$ and $\psi$ are continuous. – Reveillark Jun 14 '20 at 19:10
  • @Reveillark Okay, the result is clear for a finite collection of topological space and so I ask if is true that if $\mathfrak{X}={X_i:i\in I}$ and $\mathfrak{Y}={Y_i: i\in I}$ are two collection of topological spaces and if $\mathfrak{F}={f_i: X_i\rightarrow Y_i:i\in I}$ is a collection of functions then the function $\Delta_{{\mathfrak{F}}}:\prod{i\in I}X_i\rightarrow\prod_{i\in I}Y_i$ defined trough the condiction is continuous $$\Delta_{{\mathfrak{F}}}(x)(i)=f_i(i)$$ for any $x\in\prod{i\in I}X_i$ and for any $i\in I$ is continuous iff $f_i$ is continuous for any $i\in I$. – Antonio Maria Di Mauro Jun 14 '20 at 20:21
  • @Reveillark So what can you say about? – Antonio Maria Di Mauro Jun 14 '20 at 20:21
  • @AntonioMariaDiMauro Both of the proofs I gave above generalize to the infinite case with minimal adjustments – Reveillark Jun 14 '20 at 20:50
  • @Reveillark Perhaps I understood: read below. – Antonio Maria Di Mauro Jun 14 '20 at 21:08
  • @Reveillark Is it correct? – Antonio Maria Di Mauro Jun 14 '20 at 21:18
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Definition

If $X$ is a topological space and if $\mathfrak{Y}=\{Y_j:j\in J\}$ is a collection of topological spaces and if $\mathfrak{F}=\{f_j:X\rightarrow Y_j:j\in J\}$ is a collection of functions then the diagonal product $\Delta_{_{\mathfrak{F}}}:X\rightarrow\prod_{j\in J}Y_j$ is a function defined through the condition $$ 1.\quad\big[\Delta_{_{\mathfrak{F}}}(x)\big](j)=f_j(x) $$ for any $x\in X$ and for any $j\in J$.

Lemma

Each function of $\mathfrak{F}$ is continuous if and only if $\Delta_{_{\mathfrak{F}}}$ is continuous.

Proof. By the condition $1$ we observe that for any $j\in J$ if $\pi_j$ is the projection $j$-th projection of $\prod_{j\in J}Y_j$ then $$ \pi_j\circ\Delta_{_{\mathfrak{F}}}=f_j $$ so that the lemma follows directely by the universal mapping theorem for products.

Now if $\mathfrak{X}=\{X_j:j\in J\}$ and $\mathfrak{Y}=\{Y_j:j\in J\}$ are a collections of topological spaces and if $\Phi=\{\phi_j: X_j\rightarrow Y_j\}$ is a collection of functions then for any $j\in J$ we define $$ f_j:=\phi_j\circ\overline{\pi_j} $$ where $\overline{\pi_j}$ is the $j$-th projection of $\prod_{j\in J}X_j$ and so we even define $$ X:=\prod_{j\in J}X_j $$ and $$ \mathfrak{F}:=\{f_j:j\in J\} $$ so that $$ \underline{\pi_j}\circ\Delta_{_{\mathfrak{F}}}=f_j=\phi_j\circ\overline{\pi_j} $$ where $\underline{\pi}_j$ is the $j$-th projection in $\prod_{j\in J}Y_j$.

So in the case where $J=\{1,2\}$ it is easy to verfy that $\Delta_{_\mathfrak{F}}$ is equal to the function above defined in the question.

Antonio Maria Di Mauro
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My answer here formulates the universal property of products as follows:

| If $f: X \to \prod_{i \in I} X_i$ is a function into a product, then $f$ is continuous iff $\forall i \in I: \pi_i \circ f: X \to X_i$ is continuous.

Here we have $\Delta: A \times B \to C \times D$ and $\pi_C \circ \Delta = f \circ \pi_A$ which is a continuous map as a composition of the continuous $\pi_A$ and $f$. Similarly, $\pi_D \times \Delta = g \circ \pi_B$ is continuous, and so $\Delta$ is continuous. Your own answer overcomplicates things: it's a straightforward application of the aforementioned universal mapping principle applied to $X_1 \times X_2 = C \times D$ (and also $X=A \times B$, but that product structure is irrelevant, except I need the continuous projections on it).

I don't see the added value of using a pure category-theoretical approach here either, BTW.

The map $\Delta$ is usually called $\prod_i f_i$ in general or $f \times g$ in this case. I think it's clearer. I use $\nabla_i f_i: X \to \prod_i X_i$ for a situation where I have maps $X \to X_i$ for all $i$ (so common domain) and we define $(\nabla_i f_i)(x)=(f_i(x))_i$ etc.), which is commonly called the diagonal product of the $f_i$. The $\Delta$ is "just" the categorical product of the maps.

Henno Brandsma
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