Show that the join \(X * Y\) of two nonempty spaces \(X\) and \(Y\) is simply-connected if \(X\) is path-connected.
Solution
Use the model \[ X*Y=(X\times Y\times I)/\sim \] in which at \(t=0\) the \(Y\)-coordinate is forgotten, giving the copy of \(X\), and at \(t=1\) the \(X\)-coordinate is forgotten, giving the copy of \(Y\). Denote these two ends simply by \(X\) and \(Y\).
1. The join \(X*Y\) is path connected.
Choose \(y_0\in Y\). The subspace \[ X*\{y_0\} \] is the cone \(CX\), hence is path connected and contains the whole \(X\)-end. Every point \([x,y,t]\) can be joined to its \(X\)-end point \([x,-,0]\) by varying only the coordinate \(t\) from its given value down to \(0\). Thus every point can be joined to the path-connected subspace \(X*\{y_0\}\).
2. Let\[ J=(X*Y)\setminus Y. \]Then \(J\) deformation retracts onto the \(X\)-end.
Every point of \(J\) has a representative \([x,y,t]\) with \(t<1\). Define \[ H_u([x,y,t])=[x,y,(1-u)t]. \] This is well defined and continuous for \(0\le u\le1\). At \(u=1\) all points lie at \(t=0\), where the \(Y\)-coordinate is forgotten, so the image is exactly \(X\). Points already in \(X\) remain fixed.
3. Any path in \(X*Y\) whose endpoints lie in \(J\) is homotopic relative to its endpoints to a path lying entirely in \(J\), provided \(X\) is path connected.
Let \[ \alpha:I\to X*Y \] have endpoints outside the \(Y\)-end. The set \[ F=\alpha^{-1}(Y) \] is closed. Choose finitely many parameter intervals whose interiors cover \(F\), with all their endpoints mapped into \(J\) and with each corresponding subpath contained in the region \(t>1/3\). It suffices to replace one such subpath at a time.
Let its endpoints be \[ [x_0,y_0,t_0],\qquad [x_1,y_1,t_1], \qquad t_0,t_1<1. \] On the region \(t>1/3\), projection to the \(Y\)-coordinate is continuous even at \(t=1\). Let \[ y(s) \] be the projected \(Y\)-path of the original subpath, and choose a path \[ p:I\to X \] from \(x_0\) to \(x_1\), which exists because \(X\) is path connected.
Replace the subpath by the concatenation of:
- the vertical segment from \([x_0,y_0,t_0]\) to \([x_0,y_0,1/2]\);
- the path \[ s\longmapsto[p(s),y(s),1/2]; \]
- the vertical segment from \([x_1,y_1,1/2]\) to \([x_1,y_1,t_1]\).
This replacement lies in \(J\). It is homotopic rel endpoints to the original subpath: in the truncated cone region \(t>1/3\), contract both paths toward the \(Y\)-end while tapering the contraction to zero at the two endpoints. Both then reduce to the same projected \(Y\)-path with the same endpoint fibers; reversing one contraction gives the required relative homotopy.
Performing these finitely many replacements removes every intersection with the \(Y\)-end and leaves the original endpoints fixed.
4. Every loop in \(X*Y\) is homotopic to a loop in the \(X\)-end.
Let \(\gamma:S^1\to X*Y\) be a loop. If \(\gamma(S^1)\subseteq Y\), choose \(x_0\in X\) and define \[ K(s,u)=[x_0,\gamma(s),1-u/2]. \] At \(u=0\) this is the original loop in the \(Y\)-end, since the \(X\)-coordinate is forgotten at \(t=1\); at \(u=1\) it is a loop in the slice \(t=1/2\subset J\). Thus in this case we may first replace \(\gamma\) by a homotopic loop in \(J\).
Otherwise choose the basepoint of the parametrization at a point of the loop lying in \(J\).
By 3, \(\gamma\) is homotopic to a loop contained in \(J\). Apply the deformation retraction of 2 to obtain a loop contained in \(X\).
5. Every loop in the \(X\)-end is nullhomotopic in \(X*Y\).
Fix any \(y_0\in Y\). The subjoin \[ X*\{y_0\} \] is a cone on \(X\). It contains the \(X\)-end and is contractible by moving the join coordinate from \(t=0\) to the cone point \(y_0\) at \(t=1\). Hence every loop in \(X\) contracts inside this subspace of \(X*Y\).
6. Therefore\[ \boxed{\pi_1(X*Y)=0,} \]and since \(X*Y\) is path connected, it is simply connected.
By 4 every loop is homotopic to one in \(X\), and by 5 every such loop is nullhomotopic. Path connectedness is 1.