# Why topological embedding continuous?

Problem

Why do we require a topological embedding to be continuous? Compared to other categories we want a space to be isomorphic to some subspace. Translating this to topological spaces that is saying there is a homeomorphism to a subset in the subspace topology and considering this map as with codomain of the ambient space that doesn't need to be continuous: $X\cong B\leq Y$

Idea

Is it true that for: $f:X\to Y\text{ continuous}$ and $g:Y\to X\text{ continuous}$
$f\text{ has continuous left inverse}\iff f\text{ is topological embedding}$ $g\text{ has continuous right inverse}\iff g\text{ is quotient map}$

• I don't understand the "Problem" part, a topological embedding of $X$ into $Y$ is precisely a homeomorphism of $X$ to a subspace $B$ of $Y$, so an isomorphism to a subspace of $Y$ in $\mathbf{Top}$. Regarding the second part, both are wrong, there are embeddings without continuous left inverse, and quotient maps without continuous right inverse. – Daniel Fischer Mar 22 '14 at 15:37
• Problem Part: All authors require the homeomorphism to be also continuous considered as map to the ambient space rather than the subspace Idea Part: Ok, hmm, so thats a wrong direction... another idea? – C-Star-W-Star Mar 22 '14 at 15:42
• But if $f \colon X \to Y$ is a map, where $X$ and $Y$ are topological spaces, and $B = f(X)$ is endowed with the subspace topology induced by $Y$, then $f$ is continuous if and only if $\tilde{f}\colon X \to B$ is continuous, where $\tilde{f}(x) = f(x)$. For the subspace topology, requiring that the map be continuous as a map into the subspace, or as a map into the ambient space, are one and the same thing. It's the definition of the subspace topology as the initial topology with respect to the inclusion. – Daniel Fischer Mar 22 '14 at 15:49
• Ah right so requiring it to be continuous is necessary anyway - can u put that as answer so i can close this question in my list, please – C-Star-W-Star Mar 22 '14 at 15:55

If $i:A\to(X,\tau)$ is injective, then we can equip $A$ with the initial topology for $i$, that is the family $\{i^{-1}(U)\mid U\in\tau\}$. This is the coarsest topology making $i$ continuous, and a function $f$ from a space to $A$ is continuous iff $i\circ f$ is continuous. I particular, if $A\subset X$, and $i(a)=a$, then this gives the usual subspace topology $\{U\cap A\mid U\in\tau\}$. A map $f:Y\to X$ such that $f(Y)\subseteq A$ is now continuous iff its restriction $f':Y\to A$ is continuous, as $f=i\circ f'$.
Note that if $i:A\to X$ is an embedding, and $r:X\to A$ is a retraction, then $A\cong i(A)\subseteq X$ and $ir:X\to X$ sends every point in $X$ to a point in $i(A)$ and is constant on $i(A)$. So the categorical retractions are equivalent to the retractions as they are usually introduced in topology.
An example of an embedding without a retraction is $\{0\}\times I\cup X\times\{0\}\subset X\times I$, where $X=\{0,1/n\mid n\in\Bbb N\}$.
A quotient map without a section could be the quotient map $$q:I\to\{[0,1/2],(1/2,1]\},\qquad q(x)=A\iff x\in A$$
• Perhaps a simpler example of embeddings without retractions: $S^1 \subset D^2$, any disconnected subspace of a connected space. For a quotient map without section, $t \mapsto e^{it}$, the universal covering of the circle. – Daniel Fischer Mar 22 '14 at 16:26
• @DanielFischer: Yes, $S^0\subset D^1$ is probably the simplest example. I was dealing with cofibrations recently, that's why I came up with my example, as it shows that $(X,\{0\})$ is not cofibered. – Stefan Hamcke Mar 22 '14 at 16:36