Pushouts


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Pushouts

Definition 1.7   a Pushouts or fibered co-product of a pair of functions $ f:A\rightarrow B$ and $ g:A\rightarrow C$ in a category $ \mathscr{C}$ is a pair of $ \mathscr{C}$-arrows $ h:B\rightarrow D$ and $ k:C\rightarrow D$, such that the following conditions are satisfied:
  1. $ h o f=k o g$ i.e the following diagram commutes

    $\displaystyle \xymatrix{ A\ar[rr]^{f}\ar[dd]_{g}&&B\ar[dd]^{h}\\ &&\\ C\ar[rr]_k&&D\\ }$

    One usually writes $ D=C+_A B$

  2. Given two functions $ i:B\rightarrow E$ and $ j:C\rightarrow E$, where $ i o f=j o g$ then, there exists a unique $ \mathscr{C}$-arrow l from D to E such that the outer rectangle of the following diagram commutes

    $\displaystyle \xymatrix{ A\ar[rr]^f\ar[dd]^g&&B\ar[dd]^h\ar@/^/[rrddd]^i&&\\ &&&&\\ C\ar[rr]_k\ar@/_/[rrrrd]^j&&D\ar@{-->}[rrd]^l&&\\ &&&&E\\ }$

    i.e.

    $\displaystyle i=l o h\hspace{.2in}j=l o k$    

    We then say that f (respectively g) has been pushed out along g (respectively f)

Example For example given three sets A, B, C, the set $ D=C+_A B$ always exists and it is identified with the disjoint union of A and B i.e.

$\displaystyle D=C\amalg B:=\{(x,t)\in (C\cup B)\times\{0,1\}\vert\begin{cases}x\in C& if\hspace{.1in}t=0\\ x\in B& if\hspace{.1in}t=1 \end{cases}\}$    

where in this case the arrows h and k are defined as follows:
$ h:B\rightarrow C\amalg B$, ( $ b\longmapsto (b,0)$) and $ h:C\rightarrow C\amalg B$, ( $ c\longmapsto (c,1)$).
We now want to prove that $ C\amalg B$ as defined does indeed satisfy the conditions of a pushout.

Proof. :
Given a set E and $ j:C\rightarrow E$, $ i:B\rightarrow E$, we define the map $ l:C\amalg B\rightarrow E$ such that $ (c,0)\longmapsto j(c)$ and $ (b,1)\longmapsto i(b)$ as required form the definition of $ C\amalg B$.
It is then easy to see that the diagram

$\displaystyle \xymatrix{ A\ar[rr]^f\ar[dd]^g&&B\ar[dd]^h\ar@/^/[rrddd]^i&&\\ &&&&\\ C\ar[rr]_k\ar@/_/[rrrrd]^j&&D\ar@{-->}[rrd]^l&&\\ &&&&E\\ }$

commutes. In fact we have the following:
$ (l\circ h)(b)=l(b,0)=i(b)$ and $ (l\circ k)(b)=l(c,1)=j(c)$.

The second step in the prof is showing that the map l is unique. In fact given another map $ m:C\amalg B\rightarrow E$ such that $ m\circ h=i$ and $ m\circ k=j$, then we would have the following equality:
$ l(b,0)=i(b)=m(h(b))=m(b,0)$ and $ l(c,1)=j(c)=m(k(c))=j(c)$.
This shows that l is unique. $ \qedsymbol$


Next: Exponentiation Up: Topos Previous: Examples of Pullback
Cecilia Flori 2007-02-04