Is this true: Graph is a forest exactly when a graph with $n$ vertices and $c$ components will have exactly $n-c$ edges?
We have shown the following statement: A forest on $n$ vertices with $c$ connected components has exactly $n-c$ edges. So, I am curious, if this theorem is true in both ways?
Yes, this is true. We’ll prove that a graph $\Gamma$ with $n\in \mathbb N ^+$ vertices, $c \in \mathbb N^+$ components and $n-c$ edges is a forest. We’ll assume that the case for trees ($c=1$) has already been proven.
Let $\Gamma_1,\ldots,\Gamma_c$ be the connected components of $\Gamma$. Per definition $\Gamma$ is a forest if and only if all of these components are trees. For all $i\in \{1,\ldots,c\}$ we have $|E(\Gamma_i)| \geq |V(\Gamma_i)|-1$, because all components are connected. Now clearly $$n-c=|E(\Gamma)|=\sum_{i=1}^c |E(\Gamma_i)|\geq\sum_{i=1}^c(|V(\Gamma_i)|-1)=n-c,$$ making the inequality an equality, and thus we have $E(\Gamma_i)=V(\Gamma_i)-1$ for all $i$, because of the inequalities already established above. So all components are trees.