A map is continuous when pulling any open set back through it leaves an open set. No distance appears, no closeness is measured, and nothing is said about what the map does to an open set going forwards.

A window is judged by what can be seen through it, not by what is painted on it. Ask what each clear view corresponds to on your side of the glass.

Let $f$ take the points of $X$ to the points of $Y$, with topologies $\tau_X$ and $\tau_Y$. The preimage $f^{-1}(V)$ is the set of points of $X$ that $f$ sends into $V$; it makes sense for every $f$ and needs no inverse map. The map is continuous when every open set of $Y$ has open preimage.

• The condition runs backwards along the map. A continuous map may well send an open set to one that is not open.
• A constant map is continuous: the preimage of an open set is $X$ if the value is in it and $\varnothing$ otherwise, and both are open.
• Every map out of a discrete space is continuous, because every subset of a discrete space is open.
• Every map into an indiscrete space is continuous, because the only open sets to pull back are $\varnothing$ and $Y$.

Continuity is a demand made backwards. Take any open $V$ in $Y$, gather every point of $X$ that $f$ sends into it — that is the preimage $U = f^{-1}(V)$ — and $f$ is continuous exactly when $U$ comes out open, for every such $V$. Both are drawn dashed because both are open. Nothing at all is asked of $f$ in the other direction: a continuous map may send an open set to one that is not open.

Take $f$ from $X$ to $Y$ and $g$ from $Y$ to $Z$, both continuous, and take an open set of $Z$. Pull it back through $g$ and continuity of $g$ leaves an open set of $Y$. Pull that back through $f$ and continuity of $f$ leaves an open set of $X$. The two pullbacks done one after the other are exactly the pullback through the composite, because a point is sent into the open set by the composite precisely when $f$ sends it into the first pullback. So the composite satisfies the definition.

Take any open set of the last space; the condition has to be checked for it.

The second map is continuous, so this preimage is open in the middle space.

The first map is continuous, so the preimage of that open set is open in the first space.

A point is sent into $W$ by the composite exactly when $f$ sends it into $g^{-1}(W)$, so the two sets are the same one and the composite is continuous.

Defined for every map, invertible or not.

Only two possible preimages, and both are open.

Being a subset is already enough to be open here.

One pullback expressed as two.