Does two homotopic map induce the same homomorphism of homotopy groups? Assume $f,g$ are two maps from space $F$ to $E$ homotopic, then will they induce the same homomorphism of homotopy groups?
Or much stronger, assume $f$ is an inclusion from $F$ to $E$, $F$ the fiber of a fiber bundle $E$, $g$ is a constant map, then why $f$ induces $0$ homomorphism of homotopy groups? It is exercise 31 of Hatcher's Algebraic Topology, in section 4.2.
 A: I believe the most you can say is that they induce the same map on conjugacy classes for $\Pi_1(X)$ and more generally the action of $\Pi_1(X)$ on $\Pi_n(X)$.
Non-basepointed maps $[S^n , X]$ are in natural bijection with equivalence classes of elements of $\Pi_n (X)$ under the action of $\Pi_1(X)$ (Hatcher chapter 4). If $f,g:X \rightarrow Y$ are homotopic then $f,g$ induce the same map $[S^n, X]\rightarrow [S^n,Y]$. Interpreting this in the manner just mentioned you arrive at the result.
A: If "homotopic" means "homotopic through basepoint preserving maps", then the answer is obviously yes. The general case is covered by the result in Connor Malin's answer:
If $X$ is path connected, then there is a bijection
$$\pi_n(X,x_0)/\pi_1(X,x_0) \to [S^n, X] .$$
The action of the fundamental group $\pi_1(X,x_0)$ on $\pi_n(X,x_0)$ is in general non-trivial. Therefore it cannot be expected that the homomorphisms $f_*,g_* : \pi_n(F,y_0) \to \pi_n(E,x_0)$ induced by freely homotopic pointed maps $f, g : (F,y_0) \to (E,x_0)$ agree.
However, if $g$ is the constant map $c(y) \equiv x_0$ then $f_* = c_* = 0$. In fact, let $f$ be freely homotopic to any constant map and $[u] \in \pi_n(F,y_0)$, $u : (S^n,s_0) \to (F,y_0)$. Then $fu : S^n \to E$ is homotopic to a constant map, thus has an extension $\phi :D^{n+1} \to E$. But then we get a basepoint preserving homotopy $H : S^n \times I \to E$ from $fu$ to $cu$ via
$$H(s,t) = \phi((1-t)s + ts_0) .$$
Note that $\lVert (1-t)s + ts_0 \lVert \le 1$, i.e. $(1-t)s + ts_0 \in D^{n+1}$. We have $H(s,0) = \phi(s) = fu(s), H(s,1) = \phi(s_0) = x_0, H(s_0,t) = \phi(s_0) = x_0$.
Therefore in $\pi_n(E,x_0)$ we have $f_*([u]) = [fu] = [cu] = 0$.
