So i recently started studying homology groups and whenever i stumbled upon its general idea or motivation, i came across computations such as
$$ \langle a,b,c\rangle / \langle c \rangle = \langle a,b \rangle $$
where each $\langle \cdot \rangle$ is a free abelian group.
Unfortunately, my algebra background isn't solid but i learned that the reason for the given equation (and others) is the first isomorphism theorem. In order to show that the quotient group $\langle a,b,c\rangle$ and $\langle a,b\rangle$ are isomorphic, i just need to find a surjective homomorphism $$\varphi: \langle a,b,c\rangle \to \langle a,b\rangle$$ whose kernel $\ker \varphi$ is $\ker \varphi = \langle c\rangle$.
My question is rather simple: How exactly does such surjective homomorphism look like?
We have $$\langle a,b,c\rangle = \{m_1a+m_2b+m_3c \mid m_1,m_2,m_3 \in \mathbb{Z}\}$$ and $$\langle a,b \rangle = \{n_1a+n_2b \mid n_1,n_2 \in \mathbb{Z}\}$$
I was thinking of something like \begin{align*} \phi \left((m_1a+m_2b)+m_3\right) &= \phi(m_1a+m_2b)+\phi(m_3c)\\ &= \phi(m_1a)+\phi(m_2b) + \phi(m_3c) \\ &= m_1\phi(a) + m_2\phi(b) + m_3\phi(c) \\ &= m_1 + m_2\end{align*}
thus the homomorphism which sends $a \mapsto 1, b \mapsto 1$ and $c \mapsto 0$ for which we have $\ker \phi = \langle c \rangle$.
Would this be a legitimate approach? Thanks for any advise!