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I'm trying to solve exercise 5.6 in Steve Awodey's "Category Theory":


Show that a category with pull-backs and products has equalizers as follows: given arrows $f, g: A \to B$, take the pullback indicated below, where $\Delta = \langle 1_B, 1_B \rangle$:

Diagram

Show that $e: E \to A$ is the equalizer of $f$ and $g$.


So I need to prove that $f \circ e = g \circ e$, and that given any $z: Z \to A$ with $f \circ z = g \circ z$ we can define a unique arrow such that $e \circ u = z$. How do I use the pullback property to do this?

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Hint

Let's call $h$: $E$ $\rightarrow$B . Now considering $B$ $\times$B from the definition know the existence of two maps $\pi_1$ and $\pi_2$ $B\times$ $B$ $\rightarrow$B such that : :

$\pi_1$$\circ <f,g>$$=$ $f$

$\pi_2$$\circ <f,g>$$=$ $g$

Knowing this we can see:

$f\circ e$$=$$\pi_1$$\circ <f,g>$$\circ e$=$\pi_1$$\circ$ $\Delta$ $\circ h$= h=$\pi_2$$\circ$ $\Delta$ $\circ h$=$\pi_2$$\circ <f,g>$$\circ e$=$g\circ e$

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To complete the Abellan's answer: If $z:Z\rightarrow A$ is a morphism such that $f\circ z=g\circ z$.

Propose, $z:Z\rightarrow A$ and $f\circ h: Z\rightarrow B$.

Remember that product's projections are mono-source. So If two morphims are equal when composited with the projections, they are equal.

$\pi_{1}\circ <f,g>\circ z=f\circ z$

$\pi_{1}\circ\Delta_{B}\circ f\circ z= 1_{B}\circ f\circ z=f\circ z$

$\pi_{2}\circ <f,g>\circ z=g\circ z$

$\pi_{2}\circ \Delta_{B}\circ f\circ z= f\circ z$

But, by hypotesis, $g\circ z= f\circ z$.

From the pullback's propierties, we have that exist unique arrow $u$ such that $e\circ u=z$.

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