Building a space with given homology groups Let $m \in \mathbb{N}$. Can we have a CW complex $X$ of dimension at most $n+1$ such that $\tilde{H_i}(X)$ is  $\mathbb{Z}/m\mathbb{Z}$ for $i =n$ and zero otherwise?
 A: To expand on the comments:
The $i$-th homology of a cell complex is defined to be $\mathrm{ker}\partial_{i} / \mathrm{im}\partial_{i+1}$ where $\partial_{i+1}$ is the boundary map from the $i+1$-th chain group to the $i$-th chain group. Geometrically, this map is the attaching map that identifies the boundary of the $i+1$ cells with points on the $i$-cells. 
For example you could identify the boundary of a $2$-cell (a disk) with points on a $1$-cell (a line segment). In practice you construct a cell complex inductively so you will have already identified the end bits of the line segment with some $0$-cells (points). Assume the zero skeleton is just one point and you attach one line segment. Then we have $S^1$ and identify the boundary of $D^2$ with it. This attaching map is a map $f: S^1 \to S^1$.
You can do this in many ways, the most obvious is the identity map. This map has degree one. The resulting space is a disk and the (reduced) homology groups of this disk are $0$ everywhere except in $i=2$ where you get $\mathbb Z$. If you take $f: S^1 \to S^1$ to be the map $t \mapsto 2t$ you wrap the boundary around twice and what you get it the real projective plane which has the homology you want in $i=2$ (check it).
See here for the degree of a map.
Now generalise to $S^n \to S^n$. 
