Let's say we have a commutative diagram as in the following picture. The functions $f, h, g$ are all bijections. Can we conclude that $k$ is also a bijection?
I need this as part of my proof, but in my opinion, we can only conclude that, since $k \circ g = h \circ f$ and $h \circ f$ is a bijection (composition of bijections is a bijection), then $k$ is surjective (and $g$ is injective, but we already know this).
Is it possible to conclude that $k$ is also injective, because this is a part of a proof in linear algebra, and our professor told us to check for ourselves that $k$ is a bijection?
$k\circ g=h\circ f$ and by assumption $f,g,h$ are bijective (equivalently invertible, which also means their inverses are bijections), so $k=h\circ f\circ g^{-1}$ is a composition of bijections, hence a bijection.