From Separate Maps to One Map
A wedge sum joins pointed spaces by identifying their basepoints. A map defined on this joined space must give the same output to the two points that have become one. The key fact is that this agreement condition is the only compatibility condition needed: continuity of the map on the wedge is equivalent to continuity on each summand together with agreement at the identified point.
A pointed space is a topological space \(X\) together with a chosen point \(x_0\in X\), called its basepoint. Let \((X,x_0)\) and \((Y,y_0)\) be pointed spaces. Give the disjoint union \(X\bigsqcup Y\) the disjoint union topology: a subset is open exactly when its intersections with \(X\) and \(Y\) are open in their respective spaces. Identify the point \(x_0\) in the first summand with \(y_0\) in the second, and make no other identifications.
The quotient topology means that \(O\subseteq X\vee Y\) is open if and only if \(q^{-1}(O)\) is open in \(X\bigsqcup Y\). Thus, for any function \(F:X\vee Y\to Z\), continuity can be tested by composing with \(q\): \(F\) is continuous if and only if \(F\circ q\) is continuous. Indeed, for each open \(V\subseteq Z\), \(q^{-1}(F^{-1}(V))=(F\circ q)^{-1}(V)\), and the quotient-topology definition says precisely that a subset of \(X\vee Y\) is open when its preimage under \(q\) is open. We will also use the disjoint union criterion: a map out of \(X\bigsqcup Y\) is continuous if and only if its restrictions to \(X\) and \(Y\) are continuous. This follows because the preimage of an open set has open intersection with each summand exactly when both restrictions are continuous.
The Continuity Criterion
Proof. Suppose first that \(F:X\vee Y\to Z\) is continuous. The maps \(i_X\) and \(i_Y\) are continuous: each is the restriction of the continuous quotient map \(q\) to one summand, and the inclusion of a summand into the disjoint union is continuous. Therefore \(f=F\circ i_X\) and \(g=F\circ i_Y\) are continuous. Since \(i_X(x_0)=i_Y(y_0)\), applying \(F\) gives \(f(x_0)=g(y_0)\).
Conversely, suppose \(f:X\to Z\) and \(g:Y\to Z\) are continuous and \(f(x_0)=g(y_0)\). Define \(h:X\bigsqcup Y\to Z\) by \(h(x)=f(x)\) on the \(X\)-summand and \(h(y)=g(y)\) on the \(Y\)-summand. The disjoint union criterion makes \(h\) continuous. The equality at the basepoints ensures that \(h\) is constant on every equivalence class used to form the wedge: it has the same value on \(x_0\) and \(y_0\), and all other classes are singletons. Hence there is a well-defined function \(F:X\vee Y\to Z\) with \(F\circ q=h\). The quotient-topology criterion gives continuity of \(F\). Every point of \(X\vee Y\) has a representative in \(X\bigsqcup Y\), so the equation \(F\circ q=h\) determines \(F\) uniquely. Its restrictions are \(f\) and \(g\), as required. \(\square\)
The target \(Z\) need not have a chosen basepoint. The condition is not that the two maps preserve basepoints, but simply that they take the source basepoints to the same point of \(Z\). If \(Z\) is pointed and both maps are required to preserve basepoints, then this agreement is automatic.
Worked Example: Two Intervals Mapped to the Real Line
Take two copies of \(I=[0,1]\), each with basepoint \(0\), and form their wedge. Use the standard topology on \(I\) and on \(\mathbb{R}\). Define maps on the two summands by \(f(t)=2t\) and \(g(t)=-3t\).
Both maps are continuous, since each is the restriction of a linear function on \(\mathbb{R}\). At the basepoints, \(f(0)=2\cdot0=0\) and \(g(0)=-3\cdot0=0\), so they agree. The theorem therefore gives a unique continuous map \(F:I\vee I\to\mathbb{R}\) with these restrictions. For a point represented by \(t\) in the first interval, its value is \(2t\); for a point represented by \(t\) in the second interval, its value is \(-3t\).
For example, at \(t=\frac14\), the first branch has value \(2\cdot\frac14=\frac12\), while the second has value \(-3\cdot\frac14=-\frac34\). At the other endpoints, the first branch gives \(2\cdot1=2\), and the second gives \(-3\cdot1=-3\). These are distinct wedge points, so different outputs are allowed.
Check: the only point represented by both summands is the wedge basepoint. Its first-branch value is \(f(0)=0\), and its second-branch value is \(g(0)=0\). Thus the formula assigns exactly one real number to every point of the wedge, including the identified point.
Worked Example: A Map from Two Circles to One Circle
Give each circle \(S^1=\{z\in\mathbb{C}:|z|=1\}\) its subspace topology from \(\mathbb{C}\), and choose \(1\) as the basepoint on each circle. Let the target also be \(S^1\), and specify the maps \(f(z)=z^2\) on the first circle and \(g(z)=\overline{z}\) on the second.
For \(z,w\in S^1\), \(|z^2-w^2|=|z-w||z+w|\le 2|z-w|\), because \(|z+w|\le|z|+|w|=2\). This inequality proves continuity of \(z\mapsto z^2\). Complex conjugation satisfies \(|\overline{z}-\overline{w}|=|z-w|\), so \(g\) is continuous as well. At the basepoint, \(f(1)=1^2=1\) and \(g(1)=\overline{1}=1\). The theorem gives a unique continuous map \(F:S^1\vee S^1\to S^1\).
To see the two branches explicitly, take \(z=e^{i\pi/3}=\frac12+i\frac{\sqrt3}{2}\). On the first circle the image is \(z^2=e^{2i\pi/3}=-\frac12+i\frac{\sqrt3}{2}\). On the second circle the image is \(\overline{z}=\frac12-i\frac{\sqrt3}{2}\). Both values lie on \(S^1\): the squared modulus of each is \(\frac14+\frac34=1\). At the common basepoint \(z=1\), both branches again give \(1\), so these different branch formulas define one continuous map on the wedge.
Check: the continuity estimate for the first branch applies to every pair \(z,w\in S^1\), and the conjugation estimate does too. The basepoint calculation verifies the compatibility condition independently of the non-basepoint computations.
Worked Example: Two Branches Mapping into the Plane
Take two copies of \(I\) with basepoint \(0\), and give the target \(\mathbb{R}^2\) its Euclidean topology. Define \(f(t)=(t,t^2)\) on the first interval and \(g(t)=(t,-t^2)\) on the second. Each coordinate is a polynomial in \(t\), so each branch is continuous. More explicitly, the coordinate functions are continuous real functions, and a map into the Euclidean product \(\mathbb{R}^2\) is continuous when both coordinate functions are continuous.
The basepoint values are \(f(0)=(0,0^2)=(0,0)\) and \(g(0)=(0,-0^2)=(0,0)\). Therefore there is a unique continuous map \(F:I\vee I\to\mathbb{R}^2\) given by these two branches. At \(t=\frac12\), the first branch gives \((\frac12,(\frac12)^2)=(\frac12,\frac14)\), and the second gives \((\frac12,-(\frac12)^2)=(\frac12,-\frac14)\). At \(t=1\), the values are \((1,1)\) and \((1,-1)\), respectively.
Check: substituting \(t=0\) into both formulas gives \((0,0)\), so the identified point has a single image. The points at \(t=\frac12\) and \(t=1\) are not identified between the two summands, so their different images cause no conflict.
Compatibility Is Necessary, Not Optional
It is tempting to define a map separately on each summand and assume the quotient will automatically combine the formulas. The quotient only combines them when the formulas give the same value at the points that are identified.
Common pitfall. On two copies of \(I\) based at \(0\), define \(f(t)=t\) on the first and \(g(t)=1+t\) on the second, with target \(\mathbb{R}\). Both branches are continuous, but \(f(0)=0\) and \(g(0)=1\). A purported map on the wedge would have to assign both \(0\) and \(1\) to the single point \(i_X(0)=i_Y(0)\). Thus these formulas do not define a function on the wedge, let alone a continuous one. For a compatible correction, keep \(f(t)=t\) and use \(g(t)=2t\); now \(f(0)=g(0)=0\), so the theorem supplies a unique continuous map.
Proof. The compositions \(r\circ f\) and \(r\circ g\) are continuous because compositions of continuous maps are continuous. Their basepoint values agree, since \(r(f(x_0))=r(g(y_0))\). On the first summand, \((r\circ F)\circ i_X=r\circ(F\circ i_X)=r\circ f\); on the second, \((r\circ F)\circ i_Y=r\circ(F\circ i_Y)=r\circ g\). The wedge-sum theorem gives a unique continuous map with these restrictions, so it must be \(r\circ F\). \(\square\)
This criterion is useful whenever a space is built by joining pointed pieces and a map is easier to specify on each piece. It separates the task into two checks: establish continuity on each summand, then compare values at every point that the construction identifies. For a wedge of two spaces there is just one identified pair, so there is just one compatibility equation.
Check Your Understanding
Use the quotient definition and the continuity criterion to justify your answers.
- For two copies of \(I\) based at \(0\), do the formulas \(f(t)=t^2\) and \(g(t)=3t^2\) define a continuous map from the wedge to \(\mathbb{R}\)? Explain which condition decides.
- Give continuous maps from two copies of \(I\) based at \(1\) into \(\mathbb{R}\) that fail to define a map on the wedge, and identify the conflicting values.
- Suppose \(f:X\to Z\) and \(g:Y\to Z\) agree at the basepoints but one branch is not continuous. Does the criterion guarantee a continuous map on \(X\vee Y\)?
- For maps \(f:S^1\to S^1\) and \(g:S^1\to S^1\) based at \(1\), what equation must hold for them to induce a map from \(S^1\vee S^1\)?
- Explain why the wedge theorem gives uniqueness once the restrictions to both summands have been specified.