If $f$ and $g$ are continuous, then their Cartesian product $f\times g$ is also continuous and this can be proven using the standard open set preimage definition, but I read somewhere that an easier proof uses the projection function. How is that done?
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What do you know about the product topology? Does it have a certain universal property? – Paul Frost Feb 05 '21 at 10:38
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Do $f$ and $g$ have a common domain or not ? – Henno Brandsma Feb 05 '21 at 10:39
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f and g have different domains, but lets just assume that f the space to itself and same for g. – willyx888 Feb 05 '21 at 10:53
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The product topology has a very useful universal property, that even characterises it among other topologies:
If $X \times Y$ has the product topology and $p_X: X \times Y \to X$ and $p_Y: X \times Y \to Y$ are its natural projections, then any function $h: Z \to X \times Y$, for any space $Z$, is continuous iff both $\pi_X \circ h: Z \to X$ and $\pi_Y \circ h: Z \to Y$ are continuous.
Now suppose $f: A \to X$ and $g: B \to Y$ are both continuous. We can define $(f \times g): A \times B\to X \times Y$ by $(f \times g)(a,b)=(f(a), g(b))$ and then $f \times g$ is also continuous. This follows by the universal property just mentioned,applied to $Z=A\times B$,after noting that
$$p_X \circ (f \times g) = f \circ p_A \text{ and } p_Y \circ(f \times g)= g \circ p_B$$ are both continuous as compositions of continuous maps. (The projections $p_A,p_B$ on $A \times B$ are also continuous by definition of the product topology on that set).
Henno Brandsma
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In this answer I explain the general theory (initial topologies, of which products are an example) behind this universal theorem, among other things... – Henno Brandsma Feb 05 '21 at 11:56