Arrows in a category


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It is convenient in Category Theory to define categorical concepts externally, i.e. by reference to connections with other categories. These connections being established by functions. Therefore we will describe categorical concepts by functions.

  • Monic arrow
    Monic arrow is the "arrow-analog" of an injective function.

    Definition 1.3   An arrow $ f:a\rightarrow b$ in a Category C is monic in C if for any parallel pair $ g:c\rightarrow a,\hspace{.2in}h:c\rightarrow a$ of arrows, the equality f o g = f o h implies that h=g, i.e f is left cancellable. Monic arrows are denoted as:

    $\displaystyle \xymatrix{ *++{a}\ar@{>->}[rr]&&b\\ }$

    We now want to show how it is possible to derive a monic function from an injective one and vice versa.

    Proof. consider an injective function $ f:a\rightarrow b$ (i.e. if f(x)=f(y) then x=y) and a pair of parallel functions $ g:c\rightarrow a,\hspace{.2in}h:c\rightarrow a$ such that

    $\displaystyle \xymatrix{ c\ar[rr]^{g}\ar[dd]_{h}&&a\ar[dd]^{f}\\ &&\\ a\ar[rr]_f&&b\\ }$

    commutes, then f o g = f o h.
    Now if

    $\displaystyle \hspace{.2in}x\in C\Longrightarrow \hspace{.2in}$ $\displaystyle f o g(x)= f o h(x)$ $\displaystyle f(g(x))=f(h(x))$

    Since f is injective it follows that g(x)=h(x), i.e f is left cancellable. Vice versa, let f be left cancellable, and consider the following diagram

    \[\begin{xy}0;/r15mm/: ,(0,0)=

    It is easy to deduce that f o g=f o h since f(x)=f(y). Since x=g(0) and y=h(0) by construction, and f is left cancellable by assumption, we get: g=h, therefore x=y for f(x)=f(y), i.e. f is injective. $ \qedsymbol$

  • Epic arrow
    Epic arrow is the "arrow-analog" of a surjective function.

    Definition 1.4   An arrow $ f:a\rightarrow b$ in a Category C is epic in C if for any parallel pair $ g:b\rightarrow c,\hspace{.2in}h:b\rightarrow c$ of arrows, the equality g o f =h o f implies that h=g, i.e. f is right cancellable. Monic arrows are denoted as:

    $\displaystyle \xymatrix{ *++{a}\ar@{->>}[rr]&&b\\ }$

    An epic is a dual1.2 of a monic
  • Iso arrow
    An iso arrow is the "arrow-analog" of a bijective function.

    Definition 1.5   A C-arrow $ f:a\rightarrow b$ is iso, or invertible in C if there is a C-arrow $ g:b\rightarrow a$ such that $ g o f=1_a$ and $ f o g=1_b$. Therefore g is the inverse of f i.e. $ g=f^{-1}$.

    Theorem 1.1   g is unique.

    Proof. Consider $ g^\lq o f=1_a$ and $ f o g^\lq =1_b$, then we have
    $ g^\lq =1_a o g^\lq = (g o f) o g^\lq = g o (f o g^\lq ) = g o 1_b = g$ $ \qedsymbol$

    An iso arrow has the following properties:
    1. An iso arrow is always monic

      Proof. consider an iso f, such that f o g = f o h ( $ f:a\rightarrow b$ and $ g,h:c\rightarrow a$) then

      $\displaystyle g$ $\displaystyle = 1_a o g = (f^{-1} o f) o g = f^{-1} o (f o g)$ $\displaystyle = f^{-1} o (f o h) = (f^{-1} o f) o h = h$

      therefore f is left cancellable $ \qedsymbol$

  • An iso arrow is always epic

    Proof. consider an is f such that g o f = h o f ( $ f:a\rightarrow b$ and $ g,h:b\rightarrow c$)

    $\displaystyle g$ $\displaystyle = g o 1_b = g o (f o f^{-1}) = (g o f) o f^{-1} = (h o f) o f^{-1}$ $\displaystyle = h o (f o f^{-1}) = h$

    therefore f is right cancellable $ \qedsymbol$

    • Note: not all arrows which are monic and epic are iso, for example: inclusion map is both monic and epic, but it is not iso, otherwise it would have an inverse and, as a set function, it would have to be a bijection, but it is not. In poset even though all functions are monic and epic, only iso is the identity map. In fact, consider a function $ f:p\rightarrow q$, this implies that $ p\leq q$ if f is an iso it implies that $ f^{-1}:q\rightarrow p$ exists, therefore $ g\leq p$, but from the antisymmetry property $ p\leq q$ and $ g\leq p$ imply that p=q, therefore $ f=1_p$ is a unique arrow.

    Next: Elements and their relations Up: Category Previous: Complex example
    Cecilia Flori 2006-11-20