weak homotopy equivalence (Whitehead theorem) and the *pseudocircle*

On wikipedia, I recently read about a highly pathological finite topological space, namely the pseudocircle $$X=\{a,b,c,d\},\;\;\; \mathcal{T}=\{\emptyset,\{a\},\{b\},\{ab\},\{a,b,c\},\{a,b,d\},X\}.$$

It is stated that the map $$\begin{array}{c r c l} f\!: &\!\!\!\mathbb{S}^1&\rightarrow &\!\!\!\!\!X\\ &(x,y)&\mapsto&a\;\;\;\;\;\text{when }x<0\\ &(x,y)&\mapsto&b\;\;\;\;\;\text{when }x>0\\ &(x,y)&\mapsto&c\;\;\;\;\;\text{when }(x,y)=(0,1)\\ &(x,y)&\mapsto&d\;\;\;\;\;\text{when }(x,y)=(0,-1)\\ \end{array}$$ is a weak homotopy equivalence which by definition means that all the homomorphisms $f_\ast\!\!:\pi_n(\mathbb{S}^1,(1,0))\rightarrow\pi_n(X,b)$ are isomorphisms $\forall n\!\in\!\mathbb{N}$. "This can be proved using the following observation. Like $\mathbb{S}^1$, $X$ is the union of two contractible open sets $\{a,b,c\}$ and $\{a,b,d\}$ whose intersection $\{a,b\}$ is also the union of two contractible open sets $\{a\}$ and $\{b\}$."

But then it states "It follows that $f$ also induces an isomorphism on singular homology and cohomology". How does this follow?

The Whitehead theorem states: if $X,Y$ are connected CW-complexes and if $f\!:X\rightarrow Y$ is a weak homotopy equivalence, then it is a homotopy equivalence.

But the pseudocircle is not a CW-complex, since it only has finitely many points and isn't discrete.

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Your map $f_*$ should go the other way. – wckronholm Aug 9 '11 at 16:25
Also, the psuedocircle is not "highly pathological" as you say. The psuedocircle is in fact an example of a Khalimsky Jordan curve in the digital plane. These digital spaces have been somewhat successful in using computers to model "discrete" representations of Euclidean topologies. – wckronholm Aug 9 '11 at 16:34
@wckronholm: thanks, corrected it. As for the "highly pathological", I copied that from wiki, but it is very interesting to hear the "other side" and that there's more to the story :). If you think something interesting/important should be added on wiki, you could expand it and I'd be more than happy to read it :). Unfortunately, there's nothing on wiki about digital spaces, so I don't even know what that is... – Leon Aug 9 '11 at 18:33
Wikipedia's justification for calling this space 'highly pathological' is that it isn't a $T_1$ space. But the same can be said of any non-discrete finite topological space. – Donkey_2009 Nov 12 '14 at 17:20