kernel and cokernel of a morphism in $Map_K$

Question

Ler $Map_K$ the category whose objects are triples $(V,W,f)$, where $V$ and $W$ are finite dimensional $K$-vector spaces and $f:V\rightarrow W$ is a $K$-linear map. A morphism from $(V,W,f)$ to $(V',W',f')$ in $Map_K$ is a pair $(h_1,h_2)$ of $K$-linear maps such that $f'\circ h_1=h_2\circ f$. If $(h'_1,h'_2)$ is a morphism from $(V',W',f')$ to $(V'',W'',f'')$, we set $(h'_1,h'_2)\circ (h_1,h_2)=(h'_1h_1,h'_2h_2)$.

I want to prove that $Map_K$ is an abelian category, for this I have to prove that each morphism in $Map_K$ admits Kernel and cokernel but I dont Know how to define this.

Answer 1

I suggest you read about functor categories since this fact is a special case of something much more general. However, to explicitly answer your question the kernel of $(h_1,h_2): (V,W,f) \to (V',W',f')$ is $(\ker(h_1),\ker(h_2)) : (\text{Ker}(h_1),\text{Ker}(h_2),\bar f) \to (V,W,f)$ where $\ker(h_1) : \text{Ker}(h_1)\to V$ and $\ker(h_2) : \text{Ker}(h_2)\to W$ are the kernels of $h_1$ and $h_2$, and $\bar f : \text{Ker}(h_1)\to \text{Ker}(h_2)$ is the unique morphism such that $\ker(h_2) \bar f = f\ker(h_1)$ which exists by the universal property of $\ker(h_2)$ since $h_2 (f \ker(h_1)) = f'h_1 \ker{h_1}=0$. The construction of cokernels is dual.

Answer 2

The general categorical fact you want here is that if $C$ is a small category and $D$ is a category with limits and colimits of some shapes, then the functor category $[C, D]$ also has limits and colimits of those shapes, and they are computed pointwise. Here $D$ is finite-dimensional vector spaces and $C$ is the "walking morphism," the category with two objects and a morphism between them.