Here is the set up for this problem.
Let $A$ and $B$ be groups, $\phi: a \to B$ a group homomorphism, $A' \subset A$ a normal subgroup where $\phi$ sends $A'$ to $1_B$. I proved there is a unique group homomorphism $\tilde{\phi}: A/A' \to B$ such that $\phi = \tilde{\phi} \circ \pi$, where $\pi: A \to A/A'$ is the canonical quotient map sending $a \mapsto \overline{a}$. I am now trying to show the following consequence.
Show that $\tilde{\phi}$ is injective if and only if the inclusion $A' \subset \mathrm{ker}(\phi)$ is an equality.
Here is my attempt.
($\Rightarrow$) Suppose that $\tilde{\phi}$ is injective. Then $\mathrm{ker}(\tilde{\phi})$ is trivial, that is, $\mathrm{ker}(\tilde{\phi}) = \{1_A\}$. As $A' \subset A$ is a subgroup, $1_A \in A'$, so $\mathrm{ker}(\phi) \subset A'$. As $A' \subset \mathrm{ker}(\phi)$ by definition, we conclude $\mathrm{ker}(\phi) = A'$.
I'm not sure if there's a way to prove this without using this preliminary result that a homomorphism is injective if and only if its kernel is trivial.
($\Leftarrow$) Suppose that $A' = \mathrm{ker}(\phi)$ and that $\tilde{\phi}(y_1) = \tilde{\phi}(y_2)$ for $y_1, y_2 \in A/A'$. We must show that $y_1 = y_2$. As $\pi$ is surjective, there exists $x_1, x_2 \in A$ such that $\pi(x_1) = y_1$ and $\pi(x_2) = y_2$. Then \begin{align*} \tilde{\phi}(y_1) & = \tilde{\phi}(\pi(x_1)) = (\tilde{\phi} \circ \pi)(x_1) = \phi(x_1) \\ \tilde{\phi}(y_2) & = \tilde{\phi}(\pi(x_2)) = (\tilde{\phi} \circ \pi)(x_2) = \phi(x_2). \end{align*} So $\phi(x_1) = \phi(x_2)$, so $x_1 \sim x_2$ under the equivalence relation by which the cosets contained in $A/A'$ are defined. Therefore, $\overline{x_1} = \overline{x_2}$, so $\pi(x_1) = \pi(x_2)$, hence $y_1 = y_2$, so $\tilde{\phi}$ is injective.
How does this proof look?
