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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

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I hope you don't mind. I cleaned up the notation. –  Daniel Rust Oct 24 '13 at 19:43

1 Answer 1

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.

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