The following is based on an exercise from the book $linear algebra and geometry" by Leung.
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Show that if the diagram of linear spaces and linear transformations
$$\begin{array}{ccccccccc} X' & \xrightarrow{a} & X \\ b\big\downarrow & & \big\downarrow c \\ Y' & \xrightarrow{d} & Y \\ \end{array},\quad (1)$$
is commutative, then the linear transformations a and d define linear transformations $\text{ker }b \to \text{ker }c$
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To show the existence of the linear transformation, i extend the commutative diagram like below with the top edge of the extended commutative diagram to be the restriced map $a|_{\text{ker }b}:\text{ker }b \to \text{ker }c$.
$$\begin{array}{ccccccccc} \text{ker }b & \xrightarrow{a|_{\text{ker }b}} & \text{ker }c \\ i_b\big\downarrow & & \big\downarrow i_c \\ X' & \xrightarrow{a} & X \\ b\big\downarrow & & \big\downarrow c \\ Y' & \xrightarrow{d} & Y \\ \end{array},\quad (2)$$
I can then employ the universal property of the kernel to show existence. What I don' t understand is why I need to show the map $a|_{\text{ker }b}$ is well defined. I thought if I have a map $f:X\to Y$, unless the domain $X$ consists of cosets, then I really don't have to show the function is well defined. As the map $a|_{\text{ker }b}$ is only a restriction map. I don't recall when learning about restriction and extension maps, I need to show that they are well dedined. Can someone provide me with some clarification on thi smatter please.
Thank you in advance.