# morphisms on topological spaces

In the category of topological spaces:

1.) Show that a morphism is monic IFF it is injective

2.) Show that a morphism is epic IFF it is surjective

3.) Are there any morphisms that are monic and epic but not invertible? Prove.

4.) Show that every idempotent splits.

NOTE: A morphism $f: A \rightarrow A$ is idempotent if $f \circ f = f$. An idempotent splits if there exits $g$ and $h$ such that $f = hg$ and $gh = 1_{A}$

I understand that in general category theory, a monomorphism can be defined as:

$f: X \rightarrow Y$ such that for all $g_{1}, g_{2} : Z \rightarrow X$

$f(g_{1})=f(g_{2}) \Rightarrow g_{1}=g_{2}$

can I then, just say, therefore $f$ is injective, and then the reverse is simple.

Or have I over simplified the problem?

I would go the same route for 2.) I don't know how to approach 3.) and 4.)

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Are you sure that $gh=1_A$ and not the identity on some other space? – Stefan Hamcke Jun 9 '13 at 10:57
For (4), it suffices to show that $\mathbf{Top}$ has equalisers. – Zhen Lin Jun 9 '13 at 11:00
1) and 2) are equivalent to statement that the forgetful functor $U\colon\mathbf{Top}\to\mathbf{Set}$ preserves and reflects monics and epics. $U$ reflects monics and epics because it is faithful and U preserves monics and epics because it has both left adjoint and right andjoint. – Oskar Jun 9 '13 at 19:07

1.) Assume that $f$ is not injective. Can you find two distinct maps $g_1,g_2$ such that $f\circ g_1=f\circ g_2$? This would prove that $f$ is not a monic.

2.) Now assume that $f:X\to Y$ is not surjective. Here you have to find two distinct maps $g,h:Y\to Z$ into a space $Z$ of your choice such that $g$ and $h$ coincide on $f[X]$. HINT: A constant map is always continuous.

3.) Once you have shown that the monics are the injections and the epics are the surjection, you just have to find a continuous bijection that is not a homeomorphism.

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1. Yes, basically that's it. But spell it out, in both directions: If $f$ is injective, then $f\circ g_1=f\circ g_2$ means $f(g_1(z))=f(g_2(z))$ for all $z\in Z$, but then $g_1(z)=g_2(z)$ follows. For the other direction, choose $Z$ to be the one point space.
2. If $f:X\to Y$ is surjective, of course $g_1\circ f=g_2\circ f \implies g_1=g_2$. For the converse, if $f$ omits the value $y$, we can let $Z$ be the two point antidiscrete space $\{z_1,z_2\}$, and let $g_1$ be constant $z_1$ and let $g_2(y):=z_2$ but let it be $z_1$ everywhere else.
3. Take the identity from the two point discrete space to the two point antidiscrete space, for example.
4. If $f:A\to A$ is an idempotent, its range is $B\subseteq A$, then $f|_B=id$. Then let $h$ be the inclusion $B\hookrightarrow A$ and let $g:A\to B$ be the same as $f$ (it makes sense as the range of $f$ is $B$).
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Epimorphisms in $\mathbf{Top}$ are surjective. (Easy proof: the forgetful functor $\mathbf{Top} \to \mathbf{Set}$ has a right adjoint.) You're thinking of $\mathbf{Haus}$. – Zhen Lin Jun 9 '13 at 11:03
Aha, yes, indeed, I was presuming separation.. Thanks, I edit. – Berci Jun 9 '13 at 11:04