Solution:

There exists h: C→A such that $$(gf)\circ h = I_{C}$$. By hypothesis, $$g$$ is an isomorphism, so

\[(gf)\circ h = I_{C}\Longrightarrow\]

\[g^{-1}(gf)\circ h = g^{-1}I_{C}=g^{-1} \Longrightarrow\]

\[f\circ h \circ g= g^{-1}g=I_{C}\Longrightarrow\]

\[f\circ h=I_{C}.\]

We have the same script for the left inverse:

\[h\circ (gf) = I_{A}\]

From the uniqueness of $$f^{-1}$$, $$f^{-1}=(h\circ g)$$.

O post Categories – Exercise 1 apareceu primeiro em Educacional Plenus.

Solution:
The product A × B satisfies the universal property, for any object X and arrows r: X→A and s: X→B :

there existis only one arrow f: X → A × B such that $$\pi_{A}\circ f = r$$ and $$\pi_{B}\circ f = s$$, where π_A : A × B → A and π_B :A × B → B.

Let $$X=B\times A$$, $$r = \tilde{\pi}_{A}: B\times A \longrightarrow A$$ and $$s = \tilde{\pi}_{A}: B\times A \longrightarrow B$$, we obtain unique arrows f: B × A → A × B and g: A × B → B × A with the following properties:

• $$\pi_{A}\circ f = \tilde{\pi}_{A}$$;
• $$\tilde{\pi}_{A}\circ g = \pi_{A}$$.

Thus, we have $$\pi_{A}=\pi_{A}\circ (f\circ g)$$. (We have the same for  π_B). Clearly, $$I_{A\times B} = f\circ g$$, but, since there exist only the arrows f and g satisfying the property, we have no choice but $$I_{A\times B} = f\circ g$$.

Applying the same property for $$\tilde{\pi}_{A}=\tilde{\pi}_{A}\circ (g\circ f)$$, we have only one choice: $$g\circ f = I_{B\times A}$$.

O post Products in a Category – Exercise 1 apareceu primeiro em Educacional Plenus.

Solution:
Let $$V \in Obj(Vect_{K})$$. Any subspace $$U\subseteq V$$ and its projection $$\pi: V\longrightarrow V/U$$ form a universal pair.

Rephrasing: let any vector space $$W$$ and a linear transformation $$f\in Hom_{Vect_{k}}(V,W)$$ with $$U\subseteq ker(f)$$. There is a unique factoring funtion $$\varphi$$ such that $$\varphi\circ \pi = f$$.

i) Existence. Letting $$\varphi(v+U)=f(v)$$. The function φ is well-defined for any $$v\in V$$. If $$v+U=v’+U $$, we know $$v-v’\in U$$, therfore  $$f(v-v’) = 0$$. Since $$f$$ is linear, $$\varphi(v+U)=f(v)=f(v’)=\varphi(v’+U)$$.

Linearity of $$f$$ gives the same to φ: $$\varphi((v+U)+\alpha(w+U)) = \varphi((v+\alpha w)+U)=f(v+\alpha w)) = f(v)+\alpha f(w)$$.

ii) Uniquiness. The function $$\pi: v\mapsto v+ U$$ is surjective, and any surjective function in the category of sets is an epimorphism, the left cancellation rule holds:

\[\varphi\circ\pi = \varphi’\circ\pi \Longrightarrow \varphi = \varphi’.\]

O post Universal Property of Quotient Vector Space apareceu primeiro em Educacional Plenus.

Solution:
Let $$f:a\longrightarrow a’$$ be a isomorphism with inverse $$f^{-1}:a’\longrightarrow a$$. By definition of functors, we have $$F(Id_{a})=Id_{F(a)}$$ and $$F(f^{-1})\circ F(f)=F(f^{-1}\circ f)$$. Thus,

\[F(f^{-1})\circ F(f)=F(f^{-1}\circ f)=F(Id_{a}).\]

The same holds for $$F(f\circ f’)=F(Id_{a’})$$.

Therefore $$F(f):F(a)\longrightarrow F(a’)$$ is an isomorphism.

O post Functors – Exercise 1 apareceu primeiro em Educacional Plenus.