Suppose that $f: X \rightarrow Y$ is a continuous map and let $x_0 \in X$. Then the induced map $f_*: \pi_1(X,x_0) \rightarrow \pi_1(Y, f(x_0))$ is defined as follows. Let $[\alpha] \in \pi_1(X,x_0)$, i.e. $\alpha: I \rightarrow X$ is continuous and $\alpha(0) = x_0 = \alpha(1)$. We define
$f_*[\alpha] = [f \circ \alpha]$
Note that $f \circ \alpha:I \rightarrow Y$ is continuous and $(f \circ \alpha)(0) = f(x_0) = (f \circ \alpha)(1)$ so $f \circ \alpha$ is a loop in $Y$ at $f(x_0)$. Moreover, $f_*[\alpha]$ is well-defined: if $[\alpha] = [\beta]$ then there is an homotopy $H$ from $\alpha$ to $\beta$ and one readily verifies that $f \circ H$ is a homotopy from $f \circ \alpha$ to $f \circ \beta$.
We now show that $f_*$ is a homomorphism. The neutral element of $\pi_1(X,x_0)$ is $[c_{x_0}]$ where $c_{x_0}$ is the constant loop with value $x_0$. Then $f_*[c_{x_0}] = [f \circ c_{x_0}] = [c_{f(x_0)}]$ which is indeed the neutral element of $\pi_1(Y,f(x_0))$. We still need to show that $f_*$ preserves products, i.e. $f_*[\alpha] f_*[\beta] = f_*([\alpha][\beta])$ for any $[\alpha],[\beta] \in \pi_1(X,x_0)$.
Now, one the one hand $f_*[\alpha] f_*[\beta] = [f \circ \alpha][f \circ \beta] = [(f\circ\alpha)(f\circ\beta)]$ where
$(f\circ\alpha)(f\circ\beta)(s) = \begin{cases} (f \circ \alpha)(2s) &\mbox{if } 0 \leq s \leq \frac{1}{2} \\(f \circ \beta)(2s-1) &\mbox{if } \frac{1}{2} \leq s \leq 1 \end{cases}$
On the other hand, $f_*([\alpha][\beta]) = f_*[\alpha \beta] = [f \circ (\alpha \beta)]$ where
$(\alpha \beta)(s) = \begin{cases} \alpha(2s) &\mbox{if } 0 \leq s \leq \frac{1}{2} \\ \beta(2s-1) &\mbox{if } \frac{1}{2} \leq s \leq 1 \end{cases}$
so
$(f \circ (\alpha \beta))(s) = \begin{cases} (f \circ \alpha)(2s) &\mbox{if } 0 \leq s \leq \frac{1}{2} \\(f \circ \beta)(2s-1) &\mbox{if } \frac{1}{2} \leq s \leq 1 \end{cases}$
Therefore, $f_*[\alpha] f_*[\beta] = f_*([\alpha][\beta])$ for any $[\alpha],[\beta] \in \pi_1(X,x_0)$. So, $f_*$ is a homomorphism.
Fancy: $\pi_1(X)$ finite implies $H_1(X)$ finite implies $0 = Hom(H_1(X), \mathbb Z) = H^1(X, \mathbb Z) = [X, K(\mathbb Z, 1)] = [X, S^1]$.
Best Answer
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.