Proving that the kernel of a homomorphism is exactly equal to a normal subgroup $A'$

Here is the background for this problem.

Let $\phi: A \to B$ be a group homomorphism, $A' \subset A$ a normal subgroup where $A' \subset \mathrm{ker}(\phi)$, and $\pi: A \to A/A'$ the canonical quotient map. There exists a unique group homomorphism $\tilde{\phi}: A/A' \to B$ such that $\phi = \tilde{\phi} \circ \pi$.

I was already able to prove this fact. The group structure on $A/A'$ is "multiplication of cosets" and the map $\tilde{\phi}$ is defined by $\tilde{\phi}(y) = x$ where $\pi(x) = y$. I proved that this map is well-defined and satisfies all of the required properties. I am trying to prove the following corollary.

Show that $\tilde{\phi}$ is injective if and only if $A' \subset \mathrm{ker}(\phi)$ is an equality.

I'm fine with the backward direction. The forward direction I am struggling with. Here is my attempt.

Suppose $x \in \mathrm{ker}(\phi)$, so $\phi(x) = 1_B$. We need to show that $x \in A'$. (I believe this entails showing that $x = 1_A$ in which case $x \in A'$ because $A'$ is a subgroup.) Because $\tilde{\phi}$ is injective, its kernel is trivial, so $\mathrm{ker}(\tilde{\phi}) = \{\overline{1_A}\}$. Then $$ 1_B = \phi(x) = (\tilde{\phi} \circ \pi)(x) = \tilde{\phi}(\pi(x)). $$ As $\tilde{\phi}$ has a trivial kernel, $\pi(x) = \overline{1_A} = \overline{x}$, so $x \sim 1_A$ under the equivalence relation on the fibres of $\phi$, so $\phi(x) = \phi(1_A) = 1_B$.

This produced a statement I already knew to be true, however, as $\phi$ was a homomorphism and therefore must carry the identity to the identity. This doesn't establish that $\phi$ is injective which, while not true, would imply the result.

I would appreciate a hint on how to proceed.

Solution 1:

If there was some $x\in Ker(\phi)\setminus A'$ then we would have $\tilde{\phi}(xA')=\phi(x)=e_B$, while $xA'\ne A'$. This contradicts the assumption that the kernel of $\tilde{\phi}$ is trivial.