Let $\sigma: R \to S$ be a ring homomorphism and $\mathfrak{m}$ a maximal ideal of $S$. Show that $\sigma$ induces an injective ring homomorphism $R/\sigma^{-1}(\mathfrak{m}) \to S/\mathfrak{m}$.
Is it right to consider the canonical maps $\pi :R \to R/\sigma^{-1}(\mathfrak{m})$, $\tau: S \to S/\mathfrak{m}$ and then defining $$\varphi:R/\sigma^{-1}(\mathfrak{m}) \to S/\mathfrak{m}, \text{ as } \varphi:=\tau \circ \sigma \circ \pi^{-1}$$
this would take $$a + \sigma^{-1}(\mathfrak{m}) \longmapsto a \longmapsto \sigma(a) \longmapsto \sigma(a) + \mathfrak{m}?$$
All of the maps are ring homomorphisms so the resulting map would also be a ring homomoprhism, but I'm not entirely sure I'll get an injective map? Also I haven't used the maximality here at all so I think I'm in the wrong direction. The maximality would guarantee me that $R/\sigma^{-1}(\mathfrak{m})$ is a field?
You don't need $m$ to be maximal, this is true for any ideal. If $\sigma(a)+m=m$ then $\sigma(a)\in m$, and so by definition $a\in\sigma^{-1}(m)$, i.e $a+\sigma^{-1}(m)=\sigma^{-1}(m)$. So the kernel of your map is trivial.
Also, no, $R/\sigma^{-1}(m)$ might not be a field. For example, consider the inclusion homomorphism $\mathbb{Z}\to\mathbb{Q}$. The inverse image of the maximal ideal $\{0\}\subseteq\mathbb{Q}$ is not maximal in $\mathbb{Z}$.