A map into a product is a pair of maps, one into each factor. The theorem of this chapter says the obvious thing is true: such a map is continuous exactly when both of its coordinates are. It is the reason the product topology is the one chosen, and it makes continuity of a map into the plane a question about two functions of one variable.

A machine that outputs a position needs a dial for each axis. It is reliable exactly when both dials are, and neither can be judged by watching the other.

Checking a map into a product: the preimage of a basic box is $f_1^{-1}(U)\cap f_2^{-1}(V)$, an intersection of two preimages. It is open exactly when both coordinates pull open sets back to open sets, which is why continuity of $f$ is continuity of $f_1$ and $f_2$.

A map $f:Z\to X\times Y$ is the same thing as a pair of maps $f_1=\pi_1\circ f$ and $f_2=\pi_2\circ f$. The theorem below says $f$ is continuous exactly when $f_1$ and $f_2$ are. Nothing similar holds for maps out of a product, where being continuous in each variable separately is strictly weaker than being continuous.

• A path in the plane, $t\mapsto(x(t),y(t))$, is continuous exactly when both coordinate functions are. The same holds for a map into $\mathbb{R}^n$.
• The diagonal map $x\mapsto(x,x)$ is continuous, because both its coordinates are the identity.
• Out of a product the statement fails: $f(x,y)=\dfrac{xy}{x^2+y^2}$ with is continuous in each variable separately and not continuous at the origin, where approaching along gives .
• The theorem is a consequence of the definition by a subbasis: preimages of the subbasic strips are $f_1^{-1}(U)$ and $f_2^{-1}(V)$.
• For a product of finitely many factors the same statement holds, one condition per coordinate.

One direction is free: a composition of continuous maps is continuous, and each coordinate is the composition of the map with a projection. The other uses the subbasis test. The preimage of a subbasic strip under the map is the preimage of an open set under the corresponding coordinate, which is open by hypothesis. Since the preimages of the subbasic sets are open, the map is continuous.

Each coordinate is a composition with a projection, and projections are continuous.

For the converse, pull a subbasic strip back: the two preimages compose into one.

Both are open, because the coordinates were assumed continuous.

The preimages of all subbasic sets are open, which the subbasis test says is enough.

A map into a product is continuous exactly when each coordinate map is.

• A map into a product is continuous exactly when each coordinate is.
• The proof is the subbasis test: a strip pulls back to a coordinate preimage.
• Two continuous maps out of one space combine into a continuous map into the product.
• Out of a product, separate continuity in each variable is strictly weaker.

Great work! You now know how to:

• check continuity of a map into a product one coordinate at a time;
• combine two continuous maps into one into a product;
• prove the theorem from the subbasis test;
• give a map out of a product that is separately continuous and not continuous.