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Let $C$ be a category with an initial object $0$, pushouts, and coproducts. If $G$ and $H$ are two objects, I want to know whether $G\sqcup H$ and $G\sqcup_0 H$ are isomorphic. Letting $i_1$ and $i_2$ be the morphisms from $G$ and $H$ into $G\sqcup H$, and letting $\phi_1$ and $\phi_2$ be the morphisms from $G$ and $H$ into $G\sqcup_0 H$, I know there are two unique morphisms $i:G\sqcup_0 H\rightarrow G\sqcup H$ and $\phi:G\sqcup H\rightarrow G\sqcup_0 H$ such that $\phi_1=\phi\circ i_1$, $\phi_2=\phi\circ i_2$, $i_1=i\circ \phi_1$, and $i_2=i\circ \phi_2$.

That is, the following diagram is commutative:

$\hskip1in$ enter image description here

I believe I need to show $\phi\circ i=id_{G\sqcup H}$ and that $i\circ \phi=id_{G\sqcup_0 H}$, as they are the only morphisms showing up. But now I am stuck.

This is just for personal interest. I'd appreciate any help finishing the proof, or demonstrating a counter-example.

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The identities $\phi\circ i=1_{G\sqcup H}$ and $i\circ\phi=1_{G\sqcup_O H}$ follows from unicity of the comparison morphisms from the pushout and from $G\sqcup H$.

For example, by universal property of pushout, there exists a unique morphism $f:G\sqcup_O H\to G\sqcup_O H$ such that $f\circ \phi_1=\phi_1$ and $f\circ \phi_2=\phi_2$, and this is the identity.

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As almost always the Yoneda Lemma offers a one-line proof.

$\small \hom(G \sqcup H,K) = \hom(G,K) \times \hom(H,K) = \hom(G,K) \times_{\hom(0,K)} \hom(H,K) = \hom(G \sqcup_0 H,K)$

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  • $\begingroup$ Martin could you please justify/explain the second isomorphism and confirm whether the third formula is a pullback? $\endgroup$
    – magma
    Jul 27 '14 at 17:33
  • $\begingroup$ The second isomorphism is given by the fact that $\hom(0,K)$ is a single point. The last step is the same as the first one. $\endgroup$ Sep 6 '16 at 18:24

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