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I have a general question on using dimension formulas for linear maps.

I only know the some dimension formulas for linear maps $f:V\to W$ between $k$-vector spaces $V$ and $W$, i.e. $\dim{(V)}=\dim{(\ker(f))}+\dim {(\text{Im}(f))}$

What about if $f$ is a $\mathbb{Z}$-linear map between finitely generated abelian groups $V$ and $W$, do I have a similar dimension formular as above, i.e. $\text{rank}(V)=\text{rank}(\ker(f))+\text{rank}(\text{Im}(f))$? If yes, in which situations is it allowed to apply dimension formulas?

The background of my question is the following:

I want to calculate (for example) the simplicial homology with coefficients in $\mathbb{Z}$ for the torus $X=S^1\times S^1$ (considered as a topological space) and therefore I have to calculate the kernels and images if the simplicial boundary maps $\partial_*$. It is possible to write down transformation matrices of these boundary maps with respect to the canonical bases. Afte applying Gaussian elimination to determine the kernels and ranges I tried to apply a dimension formula as above. For the torus I obtained a correct solution, but I guess if you do this for the $\mathbb{RP}^n$ instead of the torus, again homology with coefficients in $\mathbb{Z}$, you obtain a wrong solution (if I remember correctly! I'm not 100 % sure) I guess.

Regards

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    To compute kernels and images in the chain complex, it's usually best to put the matrices representing the differentials into [Smith normal form](https://en.wikipedia.org/wiki/Smith_normal_form). – 2017-01-11

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The dimension formula for $\mathbb{Z}$-linear maps is indeed true. In fact, it can be reduced to the dimension formula for $\mathbb{R}$-linear maps by applying the tensor product over $\mathbb{R}$ (or you can work over $\mathbb{Q}$ if you prefer).

However, this is not the best way to calculate the homology of $S^1 \times S^1$ with $\mathbb{Z}$ coefficients, because it will tell you nothing about the torsion in the homology. Instead, if you have gone so far as to determine the kernels and ranges using Gaussian elimination, you should be able to go one step further and determine the quotient $\frac{\text{cycles}}{\text{boundaries}}$ precisely, not just its rank. All you are missing is a precise description of the boundaries expressed in terms of a basis for the cycles.

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    okay, thank you. Yes, I'm able to determine the quotient precisely (and I prefer other ways as Gaussian elimination, since I don't like Gaussian elimination) – 2017-01-11