In each of the following commutative diagrams assume that all maps but one are isomorphisms. Show that the remaining map must be an isomorphism as well.
\[\begin{array}{lll} A \to B & A \to B & A \to B \\ \downarrow & \downarrow & \downarrow \\ C \to D & C \to D & C \to D \end{array}\]
Solution
Hatcher’s three diagrams are respectively a commutative triangle and two commutative squares. In each case the missing map is forced to be a composite of the other maps and their inverses.
1. For the triangle\[ A\xrightarrow{f}B, \qquad A\xrightarrow{g}C\xrightarrow{h}B, \qquad f=h g, \]if any two of \(f,g,h\) are isomorphisms, then so is the third.
If \(g\) and \(h\) are isomorphisms, then \(f=hg\) is an isomorphism. If \(f\) and \(g\) are isomorphisms, then \[ h=f g^{-1}. \] If \(f\) and \(h\) are isomorphisms, then \[ g=h^{-1}f. \] In every case the remaining map is a composite of isomorphisms.
2. For the square\[ \begin{array}{ccc} A&\xrightarrow{f}&B\\ \downarrow g&&\downarrow h\\ C&\xrightarrow{k}&D, \end{array} \qquad hf=kg, \]if any three of \(f,g,h,k\) are isomorphisms, then so is the fourth.
Solve the commutativity equation for the unknown map. Explicitly, \[ f=h^{-1}kg, \qquad g=k^{-1}hf, \qquad h=kgf^{-1}, \qquad k=hfg^{-1}, \] according to which map is missing. Each expression is a composite of isomorphisms.
3. For the square with the right vertical arrow reversed,\[ \begin{array}{ccc} A&\xrightarrow{f}&B\\ \downarrow g&&\uparrow h\\ C&\xrightarrow{k}&D, \end{array} \qquad f=hkg, \]the same conclusion holds.
Again solve for the missing map: \[ f=hkg, \qquad g=k^{-1}h^{-1}f, \qquad h=fg^{-1}k^{-1}, \qquad k=h^{-1}fg^{-1}. \] Thus the fourth map is an isomorphism whenever the other three are.
Hence in each of Hatcher’s three commutative diagrams, if all maps but one are isomorphisms, the remaining map is necessarily an isomorphism as well.