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I faced a problem to understand the proof of the following theorem from the book "algebraic topology by satya deo".

If $F\colon X\to Y$ be a homotopy between two maps $ f,g\colon X\to Y $. Let $x_0\in X$ and $\sigma\colon I\to Y$ be the path joining $f(x_0)$ and $g(x_0)$ defined by $\sigma(t)=F(x_0,t)$. Then the triangle of induced homomorphisms is commutative i.e. ($\sigma_\#)\circ(f_\#)=g_\#$.

Also, if $f\colon X\to Y$ is homotopic to a constant map $C\colon X\to Y$,then the induced homomorphism $f_\#\colon\pi_1(X,x_0)\to \pi_1(Y,f(x_0))$ is the zero map.

proof:-
first part

Here I have only one problem in the first line of the theorem as author writes
Let $\alpha$ be a loop in $X$ based at $X_0$.then we know that
$$(\sigma_\#)\circ(f_\#)[\alpha]=\sigma_\#[f\circ\alpha] \\ =[\sigma^{-1}*(f\circ\alpha)*\sigma ]$$

My question is how did he write $\sigma _\#[f\circ\alpha] =[\sigma^{-1}*(f\circ \alpha) * \sigma] $ in the 2nd line instead of $[\sigma \circ f\circ\alpha]$ what he has done in the first line.

Can someone explain me please how did this happen.

second part

here he writes that "in the commutative triangle $\sigma_\#$ is isomorphism and $C_\#\colon\pi_1(X,x_0)\to \pi_1(Y,y_0)$ is a trivial map". Here both the logic I did not understand at all.

Will someone explain me elaborately please. I am new to this subject.

Thanks for your time

Dan Rust
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hutom
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2 Answers2

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Regarding the first part, it is a little unfortunate that the same notation is used for $f_\#$ and $\sigma_\#$, since those are different types of induced maps.

We have $f\colon X \to Y$, and of course $f$ induces a map of the loop spaces $\tilde{f} \colon \Omega(X,x_0)\to \Omega(Y, f(x_0))$ by compostion of a loop with $f$, $\alpha \mapsto f\circ \alpha$. That map $\tilde{f}$ induces a homomorphism of the fundamental groups $f_\# \colon \pi_1(X,x_0) \to \pi_1(Y, f(x_0))$, as you are aware.

$\sigma$ is a path in $Y$, that is, a map $I \to Y$. Of course, $\sigma$ induces in the same way a (rather boring) homomorphism $\sigma_\# \colon \pi_1(I,0) \to \pi_1(Y, f(x_0))$. But that homomorphism couldn't be composed with $f_\#$ (except if $Y = I$ and $f(x_0) = 0$), so that one cannot possibly be meant. Instead, a path in a space induces a map of the loop spaces in its start and end point, respectively, $\hat{\sigma} \colon \Omega(Y, \sigma(0)) \to \Omega(Y,\sigma(1))$, by concatenating paths $\alpha \mapsto \sigma^{-1}\ast \alpha \ast \sigma$, and that induces a homomorphism of the fundamental groups $\sigma' \colon \pi_1(Y,\sigma(0)) \to \pi_1(Y,\sigma(1))$. It is that homomorphism that is used (and also denoted by $\sigma_\#$).

Regarding the second part, $C$ is a constant map, and since $\tilde{C}$ in the notation above is also a constant map, mapping all loops to trivial (constant) loops, the induced homomorphism $C_\#$ is the trivial homomorphism. The homomorphism $\sigma'$ in my notation above, is however an isomorphism between the two fundamental groups, its inverse is induced by the reverse path $\sigma^{-1}$. So since $\sigma' \circ f_\#$ is trivial, and $\sigma'$ is an isomorphism, $f_\#$ itself must be trivial.

Daniel Fischer
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When I analysed this proof in the 1960s I found it, and the notion of change of base point, could be generalised nicely and this can be found in Chapter 7 of the book Topology and Groupoids (T&G).

The point is that the fundamental group $\pi_1(Y,y)$ is given by homotopy classes rel base point of maps $(S^1,1) \to (Y,y)$. So why not replace $(S^1,1)$, or $(S^n,1)$ for the homotopy group case, by $(X,A)$? It turns out that one needs $(X,A)$ to have the HEP, Homotopy Extension Property, to get an analogous argument. For a fixed map $u: A \to Y$ one writes $[X,Y;u]$ for the homotopy classes rel $A$ of maps $X\to Y$ which extend $u$. A homotopy $\theta: u \to v$ defines a bijection $\theta_\# : [X,Y;u] \to [X,Y;v]$. One can then prove, analogously to the arguments given in the answers on the fundamental group, that a homotopy equivalence $f: Y \to Z$ induces a bijection $$f_*: [X,Y;u] \to [X,Z;fu].$$ This result leads to a gluing theorem for homotopy equivalences, and indeed that was how I discovered it in 1965.

Ronnie Brown
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