# Understanding theorem 15.12 in Kosniovski's A first course in algebraic topology

This Theorem can be found in page 138 of the book A first Course in algebraic topology by Kosniowski.

15.12. Theorem Let $$\varphi, \psi:X \rightarrow Y$$ be continuous maps between topological spaces and let $$F$$ be an homotopy between $$\varphi$$ and $$\psi$$. Let $$f:I\rightarrow Y$$ be the path from $$\varphi(x_0)$$ to $$\psi(x_0)$$ given by $$f(t)=F(x_0,t)$$. Then, the homomorphisms

$$\varphi_* : \pi(X,x_0)\rightarrow\pi(Y,\varphi(x_0))$$ and $$\psi_* : \pi(X,x_0)\rightarrow\pi(Y,\psi(x_0))$$

are related by $$\psi_* = u_f \varphi_*$$, where $$u_f$$ is the isomorphism

$$\begin{array}{cccl} u_f: &\pi(Y,\varphi(x_0)) &\longrightarrow &\pi(Y,\psi(x_0)) \\\ &[g] &\longmapsto & [\overline{f} * g * f] \end{array}$$

I was hoping for someone to throw some light on what does this theorem say and what can it be used for.

What I understand is that the fundamental groups of the images of a point $$x_0$$ under homotopic functions are isomorphic if there is a path connecting $$\varphi(x_0)$$ and $$\psi(x_0)$$, but it has already been proven that two path connected points have isomorphic fundamental groups, so this would be redundant.

• I assume that $\pi$ should be $\pi_1$. – Arnaud Mortier Aug 17 '19 at 13:32
• @ArnaudMortier Until now the author has only talked about $\pi(X,x_0)$, the set of homotopy classes of loops with base point $x_0$ with the product $[f][g]:=[f*g]$ – Yagger Aug 17 '19 at 13:40
• That's exactly what any author I know would denote by $\pi_1(X,x_0)$. – Arnaud Mortier Aug 17 '19 at 13:44

This is essentially telling you that if two maps $$X\to Y$$ are homotopic, then they induce the "same" map at the level of homotopy groups, where the meaning of "same" here is only as broad as permitted by the fact that the two induced maps actually don't have the same codomain strictly speaking, because of the base point.
• $$X$$ is connected, $$Y$$ is not, and $$\varphi$$ and $$\psi$$ land in two different components of $$Y$$. Obviously the induced maps in homotopy can't be considered equal in any reasonable way.
• $$X=\Bbb S^1$$ with some base point that you call $$1$$, $$Y=\Bbb S^1\times \Bbb S^1$$, $$\varphi(x)=(x,1)$$ and $$\psi(x)=(1,x)$$. Here, it is interesting to see that there is an isomorphism $$u$$ from the codomain to itself such that $$\psi_* = u\circ \varphi_*$$: indeed, $$\varphi_\star =\Bbb Z\to\Bbb Z^2:n\to (n,0)$$ and $$\psi_\star =\Bbb Z\to\Bbb Z^2:n\to (0,n)$$ so that the isomorphism $$u$$ is simply $$u=\Bbb Z^2\to\Bbb Z^2:(m,n)\to (n,m)$$ But to find this isomorphism you had to understand what was going on and build it yourself. The theorem is actually telling you that when the two maps are homotopic, you don't have to think, $$u$$ is God-given.