If $m = g \circ f$ in a dagger category and $m$ is an isometry, is it possible that $f$ fails to be an isometry? Question. Suppose $m : A \rightarrow B$ is an isometry in a dagger category (by which I mean that $m^\dagger \circ m=\mathrm{id}_A$), and that we're given arrows $f : A \rightarrow Y$ and $g : Y \rightarrow B$ such that $m = g \circ f$. Is it possible that $f$ fails to be an isometry?
The remainder of the question is motivation...
In category theory, we have the following well-known result.
Proposition. Suppose $m : A \rightarrow B$ is a split monomorphism, and that we're given arrows $f : A \rightarrow Y$ and $g : Y \rightarrow B$ such that $m = g \circ f$. Then $f$ is a split monomorphism.
Proof. Since $m$ splits, let $e : B \rightarrow A$ satisfy $e \circ m = \mathrm{id}_A$. Then defining $e' = e \circ g$, we see that $$e' \circ f = (e \circ g) \circ f = e \circ m = \mathrm{id}_A.$$
Unfortunately, if we're in a dagger category and we try replacing "split monomorpism" by "isometry", the above proof doesn't seem to go through. In particular, although we can show that $f$ is a split monomorphism by defining $e' = m^\dagger \circ g$, there appears to be no guarantee that $f^\dagger$ equals $m^\dagger \circ g$.

Counterexample: let $m$ be the identity and pick $f$ and $g$ any inverses that aren't daggers of each other. For example in the dagger category of complex matrices with conjugate transpose:
$$m=\begin{pmatrix}1&0\\
0&1\\
\end{pmatrix}$$
$$f=\begin{pmatrix}1/2&0\\
0&2\\
\end{pmatrix}$$
$$g=\begin{pmatrix}2&0\\
0&1/2\\
\end{pmatrix}$$
Or even $m=1$, $f=1/2$ and $g=2$.
