Proof regarding the genus of a topological space

Question

Intuitively, I would think that if there exists a continuous mapping $f:X\rightarrow Y$ between 2 topological spaces $X$ and $Y$, then the genus of $Y$ must be smaller that the Genus of $X$ (as during the mapping, holes in $X$ can be eliminated but no new holes may be made). However, I am finding it difficult to prove this formally. I was thinking of using the transformation of loops in $X$ to $Y$ by applying $f$, but I cannot continue from this. As such, how can one prove this formally?

Answer 1

This is not true. An easy and concrete counterexample would be the quotient map (which is also the universal covering map) $f\colon \mathbb{R}^2\to \mathbb{R}^2/\mathbb{Z}^2 \cong T$ to the torus, where we identify $(x,y)\sim(\tilde x,\tilde y)$ iff $x-\tilde x \in \mathbb Z$ and $y-\tilde y \in \mathbb Z$.

This yields a continous surjective map, which is locally even a homeomorphism. But the genus of $\mathbb R^2$ is $0$, where the genus of the torus $T$ is $1$.