Homotpoic and $\#$ of two maps

Question

In general topology, and algebraic topology course i learn that

If two maps are homotopic, their $\#$ are same. $i.e$, \begin{align} f \simeq g, \quad f_{\#} = g_{\#} \end{align}

I want to know how this works.

Can you give me some proof of above statement?

The definition of $f_{\#}$ are follows If \begin{align} f: X \rightarrow Y, \qquad f_{\#} : \pi_1 (X) \rightarrow \pi_1(Y) \end{align}

Answer 1

First, let's be a little careful here : if you want to work with fundamental groups (and not groupoids), then for your question to make sense, you need to choose a base point $x_0\in X$ and assume that $f(x_0)=g(x_0)=: y_0$. Then you get maps $\pi_1(X,x_0)\to \pi_1(Y,y_0)$.

Suppose $H:I\times X\to Y$ is a homotopy between $f$ and $g$, ie $H(0,\bullet) = f$ and $H(1,\bullet) = g$ (here $I= [0;1]$).

Let $\gamma:I\to X$ be a loop based on $x_0$. Then $H\circ (Id_I \times \gamma): I\times I\to Y$ is a homotopy between $f\circ \gamma$ and $g\circ \gamma$, so $f\circ \gamma \simeq g\circ \gamma$, ie $f_{\#}([\gamma]) = g_{\#}([\gamma])$.

So indeed $f_{\#} = g_{\#}$.

Answer 2

Actually,if f and g are homotopic through H,as Lama given. Then f# and g# are not equal, actually,f# followed by r# is equal to g#.For detail,you can read Hatcher's book"algebraic topology"or see the picture "proof of Hatcher" proof of Hatcher