Show that if \(f,g:(X,\mathcal T)\to(Y,\mathcal T')\) and \(h,k:(Y,\mathcal T')\to(Z,\mathcal T'')\) are continuous and \(f\simeq g\) and \(h\simeq k\), then \(h\circ f\simeq k\circ g\). Show, however, that the converse (\(h\circ f\simeq k\circ g\) implies \(f\simeq g\) and \(h\simeq k\)) need not be true.
Solution
1. If \(f\simeq g\) and \(h\simeq k\), then \(h\circ f\simeq k\circ g\).
Let \(F\colon X\times[0,1]\to Y\) be a homotopy from \(f\) to \(g\) and \(K\colon Y\times[0,1]\to Z\) a homotopy from \(h\) to \(k\). Then \((x,t)\mapsto h(F(x,t))\) is a homotopy from \(h\circ f\) to \(h\circ g\), and \((x,t)\mapsto K(g(x),t)\) is a homotopy from \(h\circ g\) to \(k\circ g\). Homotopy is transitive.
2. With \(X=Y=S^1\), \(Z\) a point, \(f=\id_{S^1}\), \(g\) constant, and \(h=k\) the unique map \(S^1\to Z\), we have \(h\circ f=k\circ g\) but \(f\not\simeq g\).
Both composites are the unique map \(S^1\to Z\). If \(f\simeq g\), the induced homomorphisms \(\pi_1(S^1,1)\to\pi_1(S^1,f(1))\) and \(\pi_1(S^1,1)\to\pi_1(S^1,g(1))\) would agree after the change-of-basepoint isomorphism along the track of the homotopy, which for the abelian group \(\pi_1(S^1)\cong\ZZ\) does not depend on the path. But \(f_*\) is the identity of \(\ZZ\) and \(g_*=0\).
3. Q.E.D.
Step 1 proves the first assertion, and step 2 shows the converse fails.