description/proof of that for linear map between modules and linear map from supermodule of codomain of 1st map into module, composition of 2nd map after 1st map is linear
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of linear map.
- The reader knows a definition of composition of maps.
Target Context
- The reader will have a description and a proof of the proposition that for any linear map between any modules and any linear map from any supermodule of the codomain of the 1st map into any module, the composition of the 2nd map after the 1st map is linear.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules }\}\)
\(M'_2\): \(\in \{\text{ the } R \text{ modules }\}\)
\(M_2\): \(\in \{\text{ the submodules of } M'_2\}\)
\(M_3\): \(\in \{\text{ the } R \text{ modules }\}\)
\(f_1\): \(: M_1 \to M_2\), \(\in \{\text{ the linear maps }\}\)
\(f_2\): \(: M'_2 \to M_3\), \(\in \{\text{ the linear maps }\}\)
\(f_2 \circ f_1\): \(: M_1 \to M_3\)
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Statements:
\(f_2 \circ f_1 \in \{\text{ the linear maps }\}\)
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2: Note
The codomain of \(f_1\) is a submodule of the domain of \(f_2\) (not just a subset), which is the point.
3: Proof
Whole Strategy: Step 1: for each \(m, m' \in M_1\) and each \(r, r' \in R\), see that \(f_2 \circ f_1 (r m + r' m') = r f_2 \circ f_1 (m) + r' f_2 \circ f_1 (m')\).
Step 1:
Let \(m, m' \in M_1\) and \(r, r' \in R\) be any.
\(f_2 \circ f_1 (r m + r' m') = f_2 (f_1 (r m + r' m')) = f_2 (r f_1 (m) + r' f_1 (m'))\).
Note that \(r f_1 (m) + r' f_1 (m')\) is with respect to the module structure of \(M_2\).
As \(M_2\) is a submodule of \(M'_2\), \(r f_1 (m) + r' f_1 (m')\) is true also for the module structure of \(M'_2\).
So, \(f_2 (r f_1 (m) + r' f_1 (m')) = r f_2 (f_1 (m)) + r' f_2 (f_1 (m')) = r f_2 \circ f_1 (m) + r' f_2 \circ f_1 (m')\).
So, \(f_2 \circ f_1\) is linear.
