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These days, I'm working through Bredon's Topology and Geometry in order to get a better grasp on topology and fill in the gap of never having seen homology theory.

In chapter 14, Bredon proves:

If $f_0 \simeq f_1: X \to Y$ then $M_{f_0} \simeq M_{f_1} \operatorname{rel} X+Y$.

where $M_f$ is the mapping cone of $f$ (i.e. $X \times I \cup Y$ identifying $(x,0)$ with $f(x)$).

If I paraphrase, the mapping $M_{f_0} \to M_{f_1}$ is generated by "mapping the lower half of $X \times I$ to the homotopy (of $f_0 \simeq f_1$)" and similarly for $M_{f_1} \to M_{f_0}$.

The composition of these mappings should then be checked to be homotopic to the identity.

This seems easy enough, because under this mapping $X \times I$ partly maps onto a strip in $Y$, doubles up on itself, to end as the identity on the adjoining line $X \times \{0\}$. Thus one would say that we can "push" what is mapped to $Y$ back into the $X \times I$ part with ease.

Indeed, this agrees with what Bredon describes. However, he subsequently insists that the continuity of this homotopy be checked.

To me, this seems like a standard definition for a homotopy, and continuity seems glaringly obvious. Particularly in face of the standard lemma that a quotient map $X \times I \sqcup Y \to M_f$ yields a quotient map $(X \times I \sqcup Y)\times I \to M_f \times I$ (so that continuity may be checked on $X \times I^2$ and $Y\times I$).

Hence my question:

Is Bredon being formal by checking continuity, or should I be more careful in constructing homotopies because strange stuff may happen?

If the latter, an enlightening example would be highly valued.

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    $\begingroup$ The usual question arises: If you define a homotopy $$M_f\times I\to M_f$$ via a map $$X×I×I⊔Y×I \to X×I⊔Y$$ then why is this homotopy continuous, in view of the fact that $M_f×I$ is a product and not a quotient? The answer is that the product topology coincides with the quotient topology. Moreover, you have a map to a quotient, so you have to address that issue as well. $\endgroup$ – Stefan Hamcke Sep 13 '15 at 21:41
  • $\begingroup$ @StefanHamcke But indeed, this is just the usual lemma that a quotient $A \to B$ yields a quotient $A \times K \to B \times K$ if $K$ is locally compact. What is so specific about this situation that continuity needs to be made so explicit? For, the mapping to a quotient issue is already resolved by making sure that the mapping is well-defined to begin with. $\endgroup$ – Lord_Farin Sep 14 '15 at 15:11
  • $\begingroup$ The mapping to a quotient is continuous if it factors through $X×I⊔Y$. But this is a disjoint union, and while (I guess it's) $X\times[3/4,1]$ is sent to $X×I$, $X×[0,3/4]$ is sent to $Y$, at $t=0$. During the homotopy, a growing part of $X×I$ is sent to $X×I$. I think we get something like $\{(x,s,t)\mid s\ge 3(1-t)/4\}$ which is sent to $X×I$. But then again, this gives closed subsets covering $X×I×I\sqcup Y×I$, and the topology on that space is coherent with this covering, so continuity is not a big deal. But other than that, I don't know what else could cause a problem. $\endgroup$ – Stefan Hamcke Sep 14 '15 at 17:32
  • $\begingroup$ @Stefan Thanks, there were some minor points I hadn't realised but other than that your observations agree with what I noticed. 't Would be great if you'd find the time to summarise your comments in an answer :). $\endgroup$ – Lord_Farin Sep 14 '15 at 21:31

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