Let $\Sigma_n$ and $\Sigma_m$ be two closed oriented surfaces of genus $n$ and $m$, with $n \leq m$. We may think about these surfaces as connected sums of tori, so there is an canoical inclusion map $j: \Sigma_n \to \Sigma_m$.
Furthermore, we know that the fundamental group of a closed oriented surface $\Sigma_g$ of genus $g$ is given by \begin{equation} \pi_1(\Sigma_g)= \langle x_1,y_1,\ldots,x_g,y_g \mid \prod_{i=1}^g[x_i,y_i]=e \rangle \end{equation}
My question is: how does the induced map $\pi_1(j)$ look like on fundamental groups? I do have trouble visualizing how the fundamental polygon of $\Sigma_n$ lies in the fundamental polygon of $\Sigma_m$. Any help would be appreciated.
For at least "a little bit canonical" maps beween closed oriented surfaces, you need the contrary $n\geq m$. (you tried to imagine a map from a torus to $\Sigma_g$ which didn't work out)
(In the following, by component I mean the $i$-th part of the surface homeomorphic to a torus with $0,1$ or $2$ discs removed.)
A map $\Sigma_n \to \Sigma_m$ with $n \geq m$ could map some components canonically and some constant. In such a case you always get mapped the two generators of the $i$-th component canonically to the $j$-th component, and the two generators from the collapsed parts will be mapped to zero. This would correspond to a map $\mathbb Z^n \to \mathbb Z^m$ with corresponding matrix
$$(I_{m\times m} | 0 )$$
(note that this map has to look like this, in particular must have rank equal to $m$)
where you choose the basis of $\mathbb Z^n$ so that all canonical generators corresponding to a collapsed component of $\Sigma_n$ are appended in the end.