Question 2 in Chapter 1.B in Hatcher's Algebraic Topology:

Let $X$ be a connected CW complex and $G$ a group such that every homomorphism $\pi_1(X)\to G$ is trivial. Show that every map $X\to K(G, 1)$ is nullhomotopic.

Some defintions:

A $K(G,1)$ space is a path connected spaces whose fundamental group is isomorphic to a given group $G$ and which has a contractible universal covering space.

A map is nullhomotopic if is homotopic to the constant map.

A theory I have been thinking of using is Theorem 1B.9 in the text:

Let $X$ be a connected CW complex and let $Y$ be a $K(G,1)$. Then every homomorphism $\pi_1(X, x_0)\to\pi_1(Y, y_0)$ is induced by a map $(X, x_0)\to(Y, y_0)$ that is unique up to homotopy fixing $x_0$ .


Indeed, Theorem 1B.9 is a good tool here. In particular, you'll want to use the uniqueness part of the theorem. That means that any two maps $(X,x_0)\to (K(G,1),y_0)$ which induce the same map on $\pi_1$ are homotopic. What does that tell you about a map $f:(X,x_0)\to (K(G,1),y_0)$ which induces the trivial homomorphism on $\pi_1$?

The answer is hidden below.

It tells you $f$ is homotopic to the constant map $(X,x_0)\to (K(G,1),y_0)$, since the constant map also induces the trivial homomorphism on $\pi_1$. So, if there are no nontrivial homomorphisms $\pi_1(X)\to G$, this applies to every map $f:X\to Y$ (for an appropriate choice of basepoints $x_0$ and $y_0$).


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