Map Without Lift

Question

I want to show if $M$ is compact orientable surface of genus $2$, then there exists continuous map $f : M \rightarrow S^1$ with no lift to continuous map $\overline{f} : M\rightarrow \mathbb{R}$. I just know that $M$ must be homeomorphic to torus of genus $2$ and if $\overline{f}$ exists we must have $f^{\star}(\pi_1(M))\subseteq p^{\star}(\pi_1(\mathbb{R}))$ where $p : \mathbb{R}\rightarrow S^1$ is covering map, but I have no idea how to construct such function.

Answer 1

I got an answer follwoing hints from the comments. We can retract $M$ to $S^1$ by first retract $M$ to $S^1 \vee S^1$ by smashing it onto the plane, then sending the first $S^1$ to the base point and map the second $S^1$ to $S^1$ by identity. Then if the meant map is denoted by $f : M\rightarrow S^1$, we have that $f^*(\pi_1(M))$ is not zero. If $f$ had a lift, by lifting criterion we have $f^{\star}(\pi_1(M))\subseteq p^{\star}(\pi_1(\mathbb{R}))=0$. So we get a contradiction and we get the map.