We work with vector spaces over a field. We have the following commutative diagram with exact rows:
$\iota_1$ and $\rho_2$ are the canonical inclusion and projection, and they have, respectively, a canonical retraction $\rho_1$ and a canonical section $\iota_2$. The other maps are unknown. The numbers $y_1, y_2$ and $k$ are some non-negative integers. My question is the following:
Is it possible to split the upper sequence in a way compatible with the canonical splitting of the lower sequence, i.e., to find a retraction $\beta$ of $\alpha$ and a section $\delta$ of $\gamma$ such that $$ f \circ \beta = \rho_1 \circ g \quad \text{and} \quad g \circ \delta = \iota_2 \circ h\ ?$$
Note that, if such maps $\beta$ and $\delta$ exist, they are unique, since for any retraction $g'$ of $g$ and $f'$ of $f$ the following equalities must hold:
$$ \beta = f' \circ \rho_1 \circ g \quad \text{and} \quad \delta = g' \circ \iota_2 \circ h.$$
So we can start out by choosing $f'$ and $g'$ (retractions of injective maps always exist for vector spaces), defining $\beta$ and $\delta$ as in the above formulas, and checking whether they satisfy the desired properties. Let us focus exclusively on $\beta$ (the arguments for $\delta$ are analogous). It turns out that $\beta$ is automatically a retraction of $\alpha$, since $$ \beta \circ \alpha = f' \circ \rho_1 \circ g \circ \alpha = f' \circ \rho_1 \circ \iota_1 \circ f = {\rm id}_{V_1}.$$ So far, so good. But how do we know whether $f \circ \beta = \rho_1 \circ g$? This can be rewritten as $$(f \circ f') \circ (\rho_1 \circ g) = (\rho_1 \circ g).$$
But how do we know whether we can find $f'$ satisfying this property? And notice that, if we do, then every retraction $f'$ of $f$ must satisfy this too, which seems a bit puzzling.
Edit: I figured out some explicit examples where this works and where it does not. So it does not always work, but it is not clear to me whether there is a nice sufficient condition for this compatible splitting to exist.