A question concerning fundamental groups and whether a map is null-homotopic. 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?
 A: 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.
A: 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.
A: 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.
