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Is it true that if $X$ and $Y$ are topological spaces, and $f:X \rightarrow Y$ is a continuous map and the induced group homomorphism $\pi_1(f):\pi_1(X) \rightarrow \pi_1(Y)$ is the trivial homomorphism, then we have that $f$ is null-homotopic?

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    $\begingroup$ This is definitely false, e.g. take the identity map on any simply-connected space. However, you're coming right up against a much deeper problem: based on Whitehead's theorem, you might expect that if $f:X \to Y$ induces isomorphisms $\pi_n(f) : \pi_n(X) \to \pi_n(Y)$ for all $n$ (and at any basepoint), then $f$ is a (weak) homotopy equivalence -- or just a homotopy equivalence if $X$ and $Y$ are CW-complexes. [cont.] $\endgroup$ – Aaron Mazel-Gee Apr 16 '13 at 3:15
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    $\begingroup$ But this isn't known to be true, even when $X$ and $Y$ are finite CW-complexes!! In fact, this is the statement of the generating hypothesis, which has been an open question for something like half a century. Which feels like it should be rather embarassing for algebraic topology as a field, but apparently many very smart people have devoted many years to trying to prove this conjecture, and so far they've all been unsuccessful. $\endgroup$ – Aaron Mazel-Gee Apr 16 '13 at 3:16
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    $\begingroup$ @Aaron: What am I missing? If $X = T^3$ and $Y = S^2$, then the composition $T^3\rightarrow S^3\rightarrow S^2$ given by first collapsing the 2 skeleton of $T^3$ to a point, the using the Hopf map is known to be essential, and yet induce $0$ maps on all homotopy and homology groups (for trivial reasons.) Why doesn't this apply to the generating hypothesis? $\endgroup$ – Jason DeVito Apr 17 '13 at 4:46
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    $\begingroup$ @JasonDeVito: Good point! Actually, I've only heard the generating hypothesis posited stably: if $X$ and $Y$ are finite CW-spectra, then $[X,Y] \to \mbox{Hom}_{\pi_*S^0}(\pi_*X, \pi_*Y)$ is injective. In the case of your example, $T^3 = (S^1)^{\times 3}$, but stably finite products and coproducts coincide, so $\Sigma^\infty T^3 \simeq \Sigma^\infty S^1 \vee \Sigma^\infty S^1 \vee \Sigma^\infty S^1$, and so the map must be trivial by cellular approximation (i.e. since $[\Sigma^\infty S^1,\Sigma^\infty S^2]=0$). $\endgroup$ – Aaron Mazel-Gee Apr 17 '13 at 7:10
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Take $X=S^{2}$, $Y=S^{2}$, and the map $f(x)=-x$. This map has degree $-1 \neq 0$, therefore it is not nullhomotopic. However, $\pi_{1} (S^{2})$ is trivial, so the induced map will be between trivial groups, and is thus trivial.

The claim you're making is too strong because it asserts that whenever $Y$ is simply connected, then any continuous map into $Y$ is null homotopic.

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Not at all.

For example, the identity $Id_ {S^2}: S^2\to S^2 $ induces trivial homomorphism in the fundamental groups - but it is not null homotopic. In fact, this identity is the generator of the second homotopy group of the sphere, which is not trivial.

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Perhaps, I can add more examples. First of all, a generalization of my previous answer: choose a space simply connected space $X$, such that $X$ has some nontrivial homotopy group. Any covering map $p:X\to B $ is a morphism such that $p$ induces trivial morphism between the fundamental groups, but it is not null homotopic. Why? Because it induces a isomorphism between the higher homotopy groups - and, therefore, assuming that there is a nontrivial homotopy group of $X$, we get that $p$ induces a nontrivial homomorphism between homotopy groups...

Here, you might conjecture that what you said is true for spaces that aren't simply connected. But it is not true either. Let $X$ be the wedge between a sphere and a circumference. You may define a map $T: X\to S^2 $ induced by $Id_{S^2} $ and the inclusion $S^1\to S^2$. Then, compose this map with the covering map $S^2\to\mathbb{R}P^2 $. We get, thus, a map $X\to\mathbb{R}P^2 $. This map induces trivial homomorphism between fundamental groups, but it induces isomorphisms between nontrivial higher homotopy groups. And the fundamental group of both $X$ and $\mathbb{R}^2$ are nontrivial - the first one has $\mathbb{Z} $ as fundamental group and the secondo one has $\mathbb{Z} _2 $ as fundamental group.

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