Topological version of second isomorphism theorem

Question

Is there a version of the second isomorphism theorem for topological spaces? That is, given $X$ and $Y$ topological spaces, is there a condition such that $(X \cup Y)/Y$ is homeomorphic to $X/(X \cap Y)$? What about an isomorphism of homology/fundamental groups?

Answer 1

Let $Z$ be a topological space. Let $X$ and $Y$ be nonempty subspaces of $Z$.

Assumption:

The identity map from $X\cup Y$, as a subspace of $Z$, to the pushout $X\cup_{X\cap Y}Y$ in the sense of categorical notations, is a homeomorphism

This is true if, say, $X$ and $Y$ are both closed or both open in $X\cup Y$, which is true if they are both closed or both open in $Z$. If you interpret $X\cup Y$ as the categorical union then this is always true. The idea is that this assumption says maps out of $X\cup Y$ (which are constant on $Y$) are the same thing as maps out of $X$ and $Y$ (which agree on $X\cap Y$ and which are constant on $Y$) which are the same thing as maps out of $X$ (which are constant on $X\cap Y$), thereby equating the universal properties of $(X\cup Y)/Y$ and $X/(X\cap Y)$.

On a similar note, we must interpret $X/(X\cap Y)$ as the categorical quotient in the instance that $X\cap Y$ is empty i.e. we must say it is $X\sqcup\{\ast\}$ for some adjoined point $\ast$.

Then there is a canonical homeomorphism $(X\cup Y)/Y\cong X/(X\cap Y)$.

The proof:

Define $F$ to be the composite $X\hookrightarrow X\cup Y\twoheadrightarrow(X\cup Y)/Y$. Observe that $F$ is constant on $X\cap Y$ so it induces a continuous $f:X/(X\cap Y)\to(X\cup Y)/Y$ (which takes the quotient point of the domain to the quotient point of the codomain, a necessary stipulation for the case $X\cap Y=\emptyset$).

Define $G_1$ to be the quotient $X\twoheadrightarrow X/(X\cap Y)$ and $G_2$ to be the composite $Y\to\ast\hookrightarrow X/(X\cap Y)$ where the point $\ast$ is included as the quotient point. Both are continuous and agree on their common domain $X\cap Y$ so by assumption there is a continuous extension of these to some $G:X\cup Y\to X/(X\cap Y)$. Obviously $G$ is constant on $Y$, hence we have an induced continuous $g:(X\cup Y)/Y\to X/(X\cap Y)$.

It's then not too hard to check $f$ and $g$ are mutually inverse. Note by the uniqueness of the maps induced from a quotient space that it suffices to show: $$\begin{align}(X\cup Y\twoheadrightarrow(X\cup Y)/Y\overset{fg}{\longrightarrow}(X\cup Y)/Y)&=(X\cup Y\twoheadrightarrow(X\cup Y)/Y)\\(X\twoheadrightarrow X/(X\cap Y)\overset{gf}{\longrightarrow}X/(X\cap Y))&=(X\twoheadrightarrow X/(X\cap Y))\end{align}$$Which is easy if you understand how $f$ and $g$ are defined.