Continuity survives the operations that build maps out of maps. Composing two continuous maps gives a continuous one — proved in the first lesson of this chapter — and so does restricting a continuous map to a subspace, or narrowing its target to a subspace holding the image. What needs care is gluing: two continuous maps that agree where their pieces meet make a continuous whole, provided the pieces are both closed or both open.

A road built in two stretches is smooth if each stretch is smooth and they meet properly at the join. The join is the only place worth inspecting.

The pasting lemma: $X=A\cup B$ with both pieces closed, a continuous map on each, and the two agreeing on the overlap. The glued map is continuous, and the proof checks one preimage: it is the union of the two pieces' preimages, each closed in a closed piece and so closed in $X$.

Let $f:X\to Y$ and $g:Y\to Z$ be continuous, $A\subseteq X$ a subspace and $B\subseteq Y$ a subspace containing $f(X)$. Then the composite, the restriction and the corestriction below are all continuous. The pasting lemma glues: if $X=A\cup B$ with $A$ and $B$ both closed, and continuous maps on each agree on $A\cap B$, the combined map is continuous.

• The inclusion $\iota_A:A\to X$ is continuous, because $\iota_A^{-1}(U)=U\cap A$ is exactly a trace, and traces are the open sets of a subspace.
• A restriction is a composition with an inclusion, so it needs no separate proof: $f|_A=f\circ\iota_A$.
• Narrowing the target to a subspace containing the image keeps continuity: the open sets of $B$ are traces $V\cap B$, and $f^{-1}(V\cap B)=f^{-1}(V)$ once the image lies in .
• Gluing needs the pieces closed, or both open. With one piece open and the other not it can fail: on $\mathbb{R}$ take $A=(-\infty,0)$ and $B=[0,\infty)$ with the value $$ on and on ; each restriction is continuous and the glued map is not.
• The restriction of a continuous map may be continuous where the whole map is not: the map above becomes continuous as soon as it is restricted to either piece.

Use the closed-set test. The preimage of a closed set splits along the two pieces: what it meets in A is the preimage under the restriction to A, which is closed in A, and A is closed in X, so that part is closed in X. The same for B. The whole preimage is the union of these two closed sets, and a union of two closed sets is closed. The agreement on the overlap is what makes the glued map well defined in the first place.

Take a closed set of the target and use the closed-set test for continuity.

Every point of the preimage lies in A or in B, and there the map is the corresponding restriction.

Each restriction is continuous, so its preimage is closed in its own piece.

A closed subset of a closed subspace is closed in the whole space.

The preimage is a union of two closed sets, so it is closed and the map is continuous.

The inclusion of a subspace pulls an open set back to its trace, which is open in the subspace by definition.

• Compositions, restrictions and corestrictions of continuous maps are continuous.
• The inclusion of a subspace is continuous, and $f|_A=f\circ\iota_A$.
• The pasting lemma glues two continuous maps that agree on the overlap, if both pieces are closed or both open.
• With one piece open and one closed, gluing can fail.

Great work! You now know how to:

• use that compositions, restrictions and corestrictions stay continuous;
• apply the pasting lemma and check its hypotheses;
• give a gluing that fails because a piece is open;
• recognise a restriction as a composition with an inclusion.