4

Suppose $\phi, \psi: (X, x_0)\to(Y, y_0)$ are homotopic. $\exists\; H: X\times I\to Y:H(s,0)=\phi(s),\;H(s,1)=\psi(s)$. The claim is the induced homomorphisms $\phi_*, \psi_*$ on fundamental groups are equal according to this post. To prove this we have to show that $\phi\circ f$ is path homotopic to $\psi\circ f$ where $f$ is a loop at $x_0$. So we define $K:I^2\to Y$ as $K(s,t)=H(f(s),t)$ which ensures $K(s,0)=\phi\circ f(s)$ and $K(s, 1)=\psi\circ f(s)$ but to say that $K$ is a path homotopy we need $K(0,t),\; K(1, t)$ to be constant. How do we satisfy this requirement?

Eric Wofsey
  • 330,363
Jack's wasted life
  • 8,849

1 Answers1

8

In this context, when one says $\phi, \psi: (X, x_0)\to(Y, y_0)$ are homotopic, that means they are homotopic as maps $(X,x_0)\to(Y,y_0)$, not just as maps $X\to Y$. This means that every stage of the homotopy preserves the basepoint: $H(x_0,t)=y_0$ for all $t$. This guarantees that $K$ is a path homotopy.

(Without this stronger assumption on the homotopy from $\phi$ to $\psi$, it is not necessarily true that they induce the same map on fundamental groups. In general, they will differ by conjugation by the element of $\pi_1(Y,y_0)$ defined by the loop $t\mapsto H(x_0,t)$.)

Eric Wofsey
  • 330,363
  • So isn't the answer to the originial question https://math.stackexchange.com/questions/617313/if-2-spaces-are-homotopy-equivalent-then-their-fundamental-group-is-the-same/617340#617340 wrong? There's no assumption that $g\circ f(x_0)=x_0$ or homotopy between $f,g$ preserves basepoint etc in the question (although op uses these in their attempted answer). – Jack's wasted life Nov 30 '17 at 08:38
  • Yes, that proof is not quite complete. A similar argument still works, since $\phi_$ and $\psi_$ can only differ by an inner automorphism of $\pi_1(Y,y_0)$, so if one is the identity map, the other is still an isomorphism. So you get that $f_g_$ and $g_f_$ are isomorphisms, which is good enough to conclude $f_$ and $g_$ are isomorphisms (though not necessarily inverses to each other). – Eric Wofsey Nov 30 '17 at 08:52
  • @EricWofsey: does the OP's path homotopy works if we add the commutativity of $\pi_1(Y,y_0)$? (according to your last para I think so). See this post for instance – C.F.G Oct 10 '20 at 10:13
  • Related. – C.F.G Oct 10 '20 at 10:48
  • @C.F.G: Yes, that's correct. – Eric Wofsey Oct 10 '20 at 13:17