Does two homotopic map induce the same homomorphism of homotopy groups?

Question

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.

Homotopy groups require a basepoint. Induced maps on homotopy groups are only defined for basepoint preserving maps. Are $f,g$ homotopic via a homotopy keeping the basepoint fixed or freely homotopic? — Paul Frost, Feb 24 '19 at 23:32
@PaulFrost That is exactly what I want to ask. In Hatcher's book he doesn't mention any information about that, just saying what I state above. — TheWildCat, Feb 25 '19 at 04:21
Homotopic maps induce naturally isomorphic maps on fundamental groupoids. What this amounts to in terms of fundamental groups is that they induce the same map up to transfer of basepoint (which of course, has to be the case because the basepoint in the codomain may change) — Maxime Ramzi, Feb 25 '19 at 08:22
By the way, you should mention that it is exercise 31 in Hatcher. — Paul Frost, Feb 25 '19 at 11:57

score 1 · Answer 1 · answered Feb 25 '19 at 02:23

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.

Paul Frost · Answer 2 · 2019-02-25T10:41:23.773

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 $f_*([u]) = [fu] = [cu] = 0$.

Does two homotopic map induce the same homomorphism of homotopy groups?

2 Answers2