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I am trying to showed that if $X$ and $Y$ are path connected then $X \coprod_f Y$ is path connected. (The adjunction space). Let $A \subset X$ , and let $f:A \to Y$ be the attaching map.

The picture I have in mind is Image.

(Note: there is some abuse of notation in what follows, in terms of re-use of symbols for function names for different paths)

So path connectedness of $X,Y$ means for every $a,b \in X$ (or $Y$) there exists a continuous map $\phi:I \to X: \phi(0)=a, \phi(1)=b$ (or equivalent for $Y$)

Let $p: X \coprod Y \to X \coprod_f Y$ be the quotient map. Consider any two points $a,b$. If $p^{-1}(a)\cap Y = \emptyset,p^{-1}(b)\cap Y = \emptyset$ then path connectedness is clear (again, also replace $Y$ with $X$), as $X$ and $Y$ are path connected.

If $p^{-1}(a) \in A$ and $p^{-1}(b) \in A$ (I know that doesn't strictly make sense, but I think it is clear what is meant!) then path connectedness is again clear, as we can take the path given by $\phi: I \to A:\phi(0)=p^{-1}(a),\phi(1)=p^{-1}(b)$ and compose with $p$ (I guess via the maps $A \to X \to X \coprod Y \to X \coprod_f Y$) to give a path $\phi' :I \to X \coprod_f Y: \phi'(0)=a, \phi'(1)=b$

So say $p^{-1}(a)$ is in $X$, but not in $A$ and $p^{-1}(b)$ is in $A$. Then there is still a path $\phi:I \to X: \phi(0)=a, \phi(1)=b$. We can then break this into two paths $\phi_1:I \to X: \phi(0)=a, \phi(1)=t$, $\phi_2:I \to X: \phi(0)=t, \phi(1)=b$, where $(a,t) \in X$, $(t,b) \in A$, and then we can just form the path that is the composition of the two paths above (probably using the gluing lemma or something similar)

I think this argument could be made rigorous - but I also feel like there is a nice simple proof of this that I am missing

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If $X$ and $Y$ are path connected, the images of $X$ and $Y$ under the idetification map $p:X\sqcup Y\to X\sqcup_fY$ are path connected. Now, if $A$ is non-empty, then $p(X)\cap p(Y)\neq\emptyset$ so $X\sqcup_fY$ is the non-disjoint union of two path-connected subspaces. You can easily check that this implies that $X\sqcup_fY$ is then itself path-connected.

Notice the hypothesis that $A$ be non-empty is necessary!

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Thank you for the hints. But does that involve some of the work I am doing above (for example showing the images of $X$ and $Y$ under $p$ are path connected)? –  Juan S Mar 21 '11 at 1:01
    
@Qwirk: well, I am assuming that you know that the image of a path-connected set under a continuous map is path-connected... If not, well, you should certainly prove it! –  Mariano Suárez-Alvarez Mar 21 '11 at 1:07
    
Ah, yes. I was confused because $X \coprod Y$ is clearly not (necessarily) path connected - but the image of $X$ and $Y$ under $p$ are, as they themselves are path connected. –  Juan S Mar 21 '11 at 1:13
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