HaCkEd By KaSpEr511

Universal of a square

January 16th, 2010

Let $P,M_1,M_2$ and $N$ be left $R$ -modules. We say that $P$ is a universal of the square diagram

$\xymatrix{P\ar[r]^{g_1}\ar[d]_{g_2}&M_1\ar[d]^{f_1}\\M_2\ar[r]_{f_2}&N}$

if whenever we have another diagram

$\xymatrix{P'\ar[r]^{g'_1}\ar[d]_{g'_2}&M_1\ar[d]^{f_1}\\M_2\ar[r]_{f_2}&N}$

there exist a unique homomorphism $h:P'\to P$ such that the following diagram commutes

$\xymatrix@C+2em@R+2em{P' \ar@/_5pt/[ddr]_{g'_2} \ar@/^5pt/[drr]^{g'_1} \ar@{..>}[dr]^h& & \\& P \ar[d]_(0.4){g_2} \ar[r]^(0.4){g_1} & M_1 \ar[d]^{f_1}\\& M_2 \ar[r]_{f_2} & N}$

We want to show that the universal is unique of to isomorphism, i.e., if we have two universal $P$ and $P'$ then they are isomorphic. Since $P$ is a universal we have an $h:P'\to P$ that makes the above diagram commutes. Similarly, considering $P'$ as a universal, we have $h':P\to P'$ that make the diagram

$\xymatrix@C+2em@R+2em{ P\ar@/_5pt/[ddr]_{g_2} \ar@/^5pt/[drr]^{g_1} \ar@{..>}[dr]^{h'}& & \\& P '\ar[d]_(0.4){g'_2} \ar[r]^(0.4){g'_1} & M_1 \ar[d]^{f_1}\\& M_2 \ar[r]_{f_2} & N}$

commmutes. Combining the two diagrams above we have the following commutative diagram

$\xymatrix@C+2em@R+2em{P \ar@/_5pt/[ddr]_{g_2} \ar@/^5pt/[drr]^{g_1} \ar@{..>}[dr]^{h'\circ h}& & \\& P \ar[d]_(0.4){g_2} \ar[r]^(0.4){g_1} & M_1 \ar[d]^{f_1}\\& M_2 \ar[r]_{f_2} & N}$

On the other hand we have the following obvious diagram

$\xymatrix@C+2em@R+2em{P \ar@/_5pt/[ddr]_{g_2} \ar@/^5pt/[drr]^{g_1} \ar@{..>}[dr]^{h'\circ h}& & \\& P \ar[d]_(0.4){g_2} \ar[r]^(0.4){g_1} & M_1 \ar[d]^{f_1}\\& M_2 \ar[r]_{f_2} & N}$

By uniqueness, it follows that $h'\circ h = 1_P$ and by similar argument, one can show that $h\circ h'=1_{P'}$ . Therefore $h$ and $h'$ are isomorphisms and therefore $P$ is isomorphic to $P'$ .