Need help proving the the pull back diagram below gives us an isomorphism $\ker(g)\to \ker(\alpha)$?.

Question

$$\require{AMScd} \begin{CD} P@>f>>A\\ @VVgV@VV \alpha V\\ B@>\beta>>C \end{CD}$$

If $\lambda : K \to P $ and $\mu:K' \to A$ are the kernels of $g$ and $\alpha$, respectively. Then let $\phi: K \to K'$ be the induced map of the kernels.

Now the pair of maps $0:K' \to B$ and $\mu : K' \to A$ satisfy $\alpha \mu = \beta 0$ so that there is a map $h:K' \to P$ satisfying $fh=\mu$ and $gh=0$. Since $gh=0$, by the universal property of kernel, there is a unique $\psi:K' \to K$ satisfying $\lambda \psi=h$.

All that remains is to show $\phi \psi = 1_{K'}$ and $\psi \phi = 1_K$. How do you do that?

The notes that I'm reading says that $$\mu \phi \psi =f \lambda \psi = fh=\mu$$ implies $$\phi \psi = 1_{K'}$$

I don't see how the first equality in the above is true (how does $\mu \phi=f \lambda$?). And why the implication follows. As far as I can see, it is not given that $\mu$ is a monomorphism.

Edit: Nevermind the last question about why the implication is true. Kernels are monic it says ahead in the notes.

The second part is true because:

$\lambda \psi \phi = h \phi$. Since $gh\phi=0=g\lambda$ and $fh\phi=\mu\phi=f\lambda$, the uniqueness property of the pullback $P$ gives $h\phi=\lambda.$ Then $\lambda \psi \phi = h\phi = \lambda$. This implies $\psi \phi = 1_K$.

Answer 1

The map between kernels $\phi:K\to K'$ is defined as the unique map such that $\mu\phi=f\lambda$. That such a map exists follows from the universal property of kernels, because $\alpha f\lambda =\beta g\lambda = \beta 0=0$.

Answer 2

Note that the induced map $\phi\colon K\to K'$ is given by universal property of kernels, i.e., $\mu\phi = f\lambda$ is automatically valid. What you need to show is that $\alpha (f\lambda) = 0$ and then use the fact that $\mu$ is kernel of $\alpha$.

And the other part you figured yourself, $\phi\psi = 1_{K'} \iff \mu\phi\psi = \mu$ precisely because $\mu$ is mono.