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I've come up with an answer to this problem that I have precisely zero faith in, so a 'yay' or 'nay' would be enormously appreciated with regards to whether I'm an idiot or not. So here it goes:

$(X,\mathcal{T})$ is a topological space, and $\mathbb{I}= \{t \in \mathbb{R} \;\; | \;\; 0 \leq t \leq 1 \}$ has the Euclidean topology. $X \times \mathbb{I}$ is given the product topology, and for $\lambda \in \mathbb{I}$, take the function $$i_\lambda: X \longrightarrow X \times \mathbb{I}, \quad x \longmapsto (x,\lambda)$$ The image of $i_\lambda$, $\text{im}(i_\lambda)$, has the topology induced by the product topology on $X\times\mathbb{I}$. I need to show that $$\phi_\lambda: X \longrightarrow \text{im}(i_\lambda),\quad x\longmapsto i_\lambda(x)$$ is a homeomorphism.

OK, so, firstly I decided that $\text{im}(i_\lambda) = X \times \{\lambda\}$.
For injectivity, given $a,b \in X$, $\phi_\lambda(a) = \phi_\lambda(b) \Longleftrightarrow (a,\lambda)=(b,\lambda) \Longleftrightarrow a=b$.
For surjectivity, given $(a,\lambda)\in X\times \{\lambda\}$, $\exists b\in X$ such that $\phi_\lambda(b) = (a,\lambda)$, namely $b=a$.
So $\phi_\lambda$ is bijective.
Now I define $$\zeta_\lambda: X\times \{\lambda\} \longrightarrow X,\quad (x,\lambda)\longmapsto x.$$ I think that $\zeta_\lambda \circ \phi_\lambda = \text{id}_X$ and that $\phi_\lambda \circ \zeta_\lambda = \text{id}_{X\times \{\lambda\}}$, so $\zeta_\lambda = \phi_\lambda^{-1}$. Now it just remains to be shown that $\phi_\lambda$ and $\phi_\lambda^{-1}$ are continuous. To this end, I showed that the topology on $\text{im}(i_\lambda)=X\times\{\lambda\}$ is
$$\mathcal{G} = \{U\times\{\lambda\}\;|\; U \in \mathcal{T}\,\},$$ and so to show that $\phi_\lambda$ is continuous, take $A \in \mathcal{G}$, i.e. $A=U\times\{\lambda\}$ for some $U\in \mathcal{T}$. Then $$\phi_\lambda^{-1}[A] = \phi_\lambda^{-1}[U\times\{\lambda\}]=U\in\mathcal{T}.$$ Thus $\phi_\lambda$ is continuous. Then for $\phi_\lambda^{-1}$, take $K\in\mathcal{T}$; $$\left(\phi_\lambda^{-1}\right)^{-1}[K] = \phi_\lambda[K] = K\times\{\lambda\} \in \mathcal{G}.$$ Hence $\phi_\lambda^{-1}$ is continuous.
Therefore $\phi_\lambda$ is a homeomorphism.

Please be as brutally honest as you feel you need to be, as I have no delusions about my mathematical ability. If it makes no sense and I understand nothing, so be it. Thank you for your time.

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A hint for future reference: A function is always onto its image. – Asaf Karagila May 14 '11 at 15:21
up vote 1 down vote accepted

Your argument seems correct, but slightly overcomplicated.

First of all, a function is always onto its image, so an injective function is automatically a bijection with its image.

As for checking the continuity of $\phi_\lambda$ simply take an open set in $X\times\{\lambda\}$, namely $U\times\{\lambda\}$ and note that its preimage is the open set $U$ and therefore $\phi_\lambda$ is continuous.

Lastly, the product topology is the finest topology under which the projections are continuous.

Therefore $\phi_\lambda^{-1}$ is continuous by that virtue, and it would require no checking.

In case you did not learn that property of the product topology, it is still simple enough to show its continuity as the preimage of $U$ is indeed $U\times\{\lambda\}$.

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Thanks! I was sure upon posting that it was overly long and byzantine, but to be honest I'm a little bit afraid of topology, and the shorter the proof the more insecure I am in its correctness. – Hargrove May 14 '11 at 16:14
@Hargrove: It is an important step in mathematics not to be afraid of short proofs, and another very important step to be able to think critically and find the flaws in your own arguments. – Asaf Karagila May 14 '11 at 16:48

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